Last page ended with an arrow: a measure that's finite on compact sets gives a positive linear functional on $C_c(X)$, namely $f \mapsto \int_X f\,d\mu$. This page runs the arrow backwards. Hand me any positive linear functional on $C_c(X)$, and I'll hand you back a measure it integrates against. Better: the measure comes out regular, so its sets can be squeezed between compact sets and open sets. That last part is what makes this the workhorse of the whole course. Lebesgue measure, on the next page, falls right out of it.
On Locally Compact Spaces & Urysohn's Lemma, the corollary said: $\mu(K) < \infty$ for every compact $K$ exactly when $C_c(X) \subseteq L^1(\mu)$, and then $\Lambda f = \int_X f\,d\mu$ is a positive linear functional. That's the forward arrow.
Here's the backwards question. Somebody gives you a positive linear functional $\Lambda$ on $C_c(X)$. It's a black box: feed it a continuous function with compact support, it hands you a number, and it never hands you a negative number for a nonnegative function. Is there a measure $\mu$ with $\Lambda f = \int_X f\,d\mu$ for every $f$? And how would you even find it?
Two warm-ups, where you can guess the answer. Both live on $X = \mathbb{R}$.
Guess a measure $\mu$ with $\Lambda f = \int_{\mathbb{R}} f\,d\mu$ for (a) $\Lambda f = f(2)$, and (b) $\Lambda f = 3f(0) + f(1)$.
(a) Integrating against the Dirac measure $\delta_2$ just reads off the value at $2$: $\int f\,d\delta_2 = f(2)$. So $\mu = \delta_2$.
(b) Integrals add over sums of measures, so $\mu = 3\delta_0 + \delta_1$ works: $\int f\,d\mu = 3f(0) + f(1)$.
Those were easy because the functional told us where to look. A real black box doesn't. But it does let you probe: slide a narrow bump around and watch what $\Lambda$ says. Wherever the readout jumps, there's mass. Try it on a functional whose formula is hidden.
Notice what you did. A tall narrow bump that equals $1$ on a small set $K$ measured, roughly, "how much stuff is in $K$". Shrink the bump onto $K$ from above and you get the mass of $K$ exactly. That's the whole proof in one sentence. The rest of the page makes it honest.
For the rest of this section, $X$ is a locally compact Hausdorff space and $\Lambda$ is a positive linear functional on $C_c(X)$. We want a measure, so we need a size for every set. Start with open sets, where Urysohn's lemma gives us lots of functions to test with. Say $f$ is under an open set $V$ when $$f \in C_c(X), \qquad 0 \le f \le \chi_V, \qquad \operatorname{supp} f \subseteq V.$$ So $f$ takes values in $[0, 1]$ and lives on a compact piece of $V$. Every such $f$ has $\Lambda f \ge 0$, and the bigger $f$ gets, the closer it comes to "all of $V$". So define the size of $V$ as the best you can do: $$\mu(V) = \sup\{\Lambda f : f \text{ is under } V\}.$$ Watch the sup happen. The functional here puts a spread of mass along an interval and a point mass at the right end of $V$, just outside it.
The bumps fill out $V$ from inside, and $\Lambda f$ climbs to the mass of $V$. But the point mass at the edge never counts: every $f$ under $V$ has its support inside $V$, so it's $0$ at that point. Open sets don't contain their boundary, and neither does their measure.
Now every other set. Cover it with open sets and take the cheapest cover, and then pick out the sets whose measure can also be reached from inside by compact sets: $$\mu(E) = \inf\{\mu(V) : V \text{ open}, \ E \subseteq V\} \quad \text{for every } E \subseteq X,$$ $$\mathcal{M}_F = \{E \subseteq X : \mu(E) < \infty \text{ and } \mu(E) = \sup\{\mu(K) : K \subseteq E, \ K \text{ compact}\}\},$$ $$\mathcal{M} = \{E \subseteq X : E \cap K \in \mathcal{M}_F \text{ for every compact } K\}.$$ For an open $E$, the two definitions of $\mu(E)$ agree: if $V_1 \subseteq V_2$ are open, every $f$ under $V_1$ is under $V_2$, so $\mu(V_1) \le \mu(V_2)$, and the cheapest open cover of an open $E$ is $E$ itself. The same argument shows $\mu$ is monotone: $A \subseteq B$ gives $\mu(A) \le \mu(B)$. And $\mu(\emptyset) = 0$, since the only $f$ under $\emptyset$ is $0$.
So $\mu$ is defined on every subset of $X$. It won't be countably additive on all of them; $\mathcal{M}$ is where it will be. The plan: show $\mu$ is subadditive everywhere, compute it on compact sets, show it's additive on $\mathcal{M}_F$, and then show $\mathcal{M}$ is a σ-algebra. One tool first. Splitting a function into pieces that each live in one of several open sets is a job Urysohn's lemma can do.
Let $X$ be a locally compact Hausdorff space, $K \subseteq X$ compact, and $V_1, \ldots, V_n$ open with $K \subseteq V_1 \cup \cdots \cup V_n$. Then there are functions $h_1, \ldots, h_n$, each $h_i$ under $V_i$, with $$h_1(x) + \cdots + h_n(x) = 1 \quad \text{for every } x \in K.$$
Proof. Each $x \in K$ lies in some $V_i$. By the shrinking theorem applied to $\{x\} \subseteq V_i$, it has an open neighborhood $W_x$ with compact closure $\overline{W_x} \subseteq V_i$. Finitely many of these, $W_{x_1}, \ldots, W_{x_m}$, cover $K$. For each $i$, let $H_i$ be the union of those $\overline{W_{x_j}}$ that lie in $V_i$. Then $H_i$ is compact (a finite union of compact sets), $H_i \subseteq V_i$, and $K \subseteq H_1 \cup \cdots \cup H_n$.
By Urysohn's lemma there are $g_i \in C_c(X)$ with $\chi_{H_i} \le g_i \le \chi_{V_i}$ and $\operatorname{supp} g_i \subseteq V_i$. (If $H_i = \emptyset$, take $g_i = 0$.) Let $$h_1 = g_1, \quad h_2 = (1 - g_1)\,g_2, \quad \ldots, \quad h_n = (1 - g_1)(1 - g_2)\cdots(1 - g_{n-1})\,g_n.$$ Each $h_i$ is continuous, $0 \le h_i \le 1$, and $h_i(x) \neq 0$ only where $g_i(x) \neq 0$, so $\operatorname{supp} h_i \subseteq \operatorname{supp} g_i \subseteq V_i$, which is compact. So $h_i$ is under $V_i$. Multiplying out (by induction on $n$), $$h_1 + \cdots + h_n = 1 - (1 - g_1)(1 - g_2)\cdots(1 - g_n).$$ Every $x \in K$ lies in some $H_i$, where $g_i(x) = 1$, so the product is $0$ and the sum is $1$. $\blacksquare$
Here's one you can draw, on the line.
On $\mathbb{R}$, let $K = [0, 3]$, $V_1 = (-1, 2)$, and $V_2 = (1, 4)$. Let $g_1$ be $1$ on $[-\tfrac{1}{2}, \tfrac{3}{2}]$, $0$ outside $(-\tfrac{3}{4}, \tfrac{7}{4})$, and linear in between, and $g_2$ be $1$ on $[\tfrac{3}{2}, \tfrac{7}{2}]$, $0$ outside $(\tfrac{5}{4}, \tfrac{15}{4})$, and linear in between. Check that $h_1 = g_1$ and $h_2 = (1 - g_1)g_2$ form a partition of unity for $K$.
The supports are $[-\tfrac{3}{4}, \tfrac{7}{4}] \subseteq V_1$ and $\operatorname{supp} h_2 \subseteq [\tfrac{5}{4}, \tfrac{15}{4}] \subseteq V_2$, and both functions take values in $[0, 1]$. On $K$, check the sum piece by piece:
So $h_1 + h_2 = 1$ on all of $K$. Where $h_1$ ramps down, $h_2$ picks up exactly what it drops.
That's the trick in the formula: $h_1$ takes as much of the job as $g_1$ wants, $h_2$ takes what's left times $g_2$, and so on, so nothing is counted twice. It's exactly what we need to compare $\mu$ of a union with the sum of the pieces.
For any sets $E_1, E_2, \ldots \subseteq X$, $$\mu\Big(\bigcup_{i=1}^{\infty} E_i\Big) \le \sum_{i=1}^{\infty} \mu(E_i).$$
Proof. Two open sets. Let $V_1, V_2$ be open and $g$ under $V_1 \cup V_2$. Its support $K$ is compact and lies in $V_1 \cup V_2$, so the partition of unity gives $h_i$ under $V_i$ with $h_1 + h_2 = 1$ on $K$. Then $g = h_1 g + h_2 g$ (both sides are $0$ off $K$), and each $h_i g$ is under $V_i$: it's continuous, $0 \le h_i g \le 1$, and $\operatorname{supp}(h_i g) \subseteq \operatorname{supp} h_i \subseteq V_i$. So $$\Lambda g = \Lambda(h_1 g) + \Lambda(h_2 g) \le \mu(V_1) + \mu(V_2).$$ Taking the sup over $g$ gives $\mu(V_1 \cup V_2) \le \mu(V_1) + \mu(V_2)$, and by induction the same holds for finitely many open sets.
Any sets. If some $\mu(E_i) = \infty$ there is nothing to prove. Otherwise let $\varepsilon > 0$, and choose open $V_i \supseteq E_i$ with $\mu(V_i) < \mu(E_i) + \varepsilon/2^i$. Let $V = \bigcup V_i$ and let $f$ be under $V$. The compact set $\operatorname{supp} f$ is covered by the $V_i$, so it lies in $V_1 \cup \cdots \cup V_n$ for some $n$, and $f$ is under that union. Hence $$\Lambda f \le \mu(V_1 \cup \cdots \cup V_n) \le \sum_{i=1}^{n} \mu(V_i) \le \sum_{i=1}^{\infty} \mu(E_i) + \varepsilon.$$ Taking the sup over $f$, $\mu(V) \le \sum \mu(E_i) + \varepsilon$. Since $\bigcup E_i \subseteq V$, monotonicity gives $\mu(\bigcup E_i) \le \sum \mu(E_i) + \varepsilon$ for every $\varepsilon > 0$. $\blacksquare$
Subadditivity is the easy half of countable additivity. The hard half needs sets we can actually compute with, and compact sets are the ones Urysohn's lemma hugs best. Open sets got their measure from below; compact sets get theirs from above.
Every compact $K \subseteq X$ is in $\mathcal{M}_F$, and $$\mu(K) = \inf\{\Lambda f : f \in C_c(X), \ \chi_K \le f \le 1\}.$$
Proof. $\mu(K) \le \Lambda f$. Let $f \in C_c(X)$ with $\chi_K \le f \le 1$, and let $0 < \alpha < 1$. The set $V_\alpha = \{f > \alpha\}$ is open and contains $K$. If $g$ is under $V_\alpha$, then $\alpha g \le f$: on $V_\alpha$, $\alpha g \le \alpha < f$, and off $V_\alpha$, $\alpha g = 0 \le f$. Since $\Lambda$ is positive, $\Lambda(f - \alpha g) \ge 0$, so $\Lambda g \le \Lambda f / \alpha$. Taking the sup over $g$, $$\mu(K) \le \mu(V_\alpha) \le \frac{\Lambda f}{\alpha}.$$ Let $\alpha \to 1$: $\mu(K) \le \Lambda f$. Urysohn's lemma (with $V = X$) gives at least one such $f$, so $\mu(K) < \infty$. And $K$ is a compact subset of itself, so $\sup\{\mu(K') : K' \subseteq K \text{ compact}\} = \mu(K)$. So $K \in \mathcal{M}_F$.
The infimum is $\mu(K)$. Let $\varepsilon > 0$. By the definition of $\mu(K)$ there is an open $V \supseteq K$ with $\mu(V) < \mu(K) + \varepsilon$. Urysohn's lemma gives $f \in C_c(X)$ with $\chi_K \le f \le \chi_V$ and $\operatorname{supp} f \subseteq V$, so $f$ is under $V$ and $\chi_K \le f \le 1$. Then $\Lambda f \le \mu(V) < \mu(K) + \varepsilon$. $\blacksquare$
Here's the infimum from the other side: tents that are $1$ on $K$, squeezing down onto it. This time the point mass sits on the end of $K$, and a compact set does contain its endpoints, so it counts.
That's enough to compute something real. Here's the example that the whole next page grows out of.
Let $\Lambda f = \int_0^1 f(x)\,dx$ (a Riemann integral) on $C_c(\mathbb{R})$, and let $\mu$ be built from $\Lambda$ as above. Show that $\mu([s, t]) = t - s$ whenever $0 \le s < t \le 1$.
$\Lambda$ is linear, and $f \ge 0$ gives $\int_0^1 f \ge 0$, so it's positive. Let $K = [s, t]$, compact. We use Lemma 2.
From below. If $\chi_K \le f \le 1$, then $f \ge 0$ on $[0, 1]$ and $f = 1$ on $[s, t]$, so $\int_0^1 f \ge \int_s^t 1\,dx = t - s$. So $\mu(K) \ge t - s$.
From above. For $\delta > 0$, let $f_\delta$ be the tent that is $1$ on $[s, t]$, $0$ outside $[s - \delta, t + \delta]$, and linear in between. Then $\chi_K \le f_\delta \le 1$, and $f_\delta$ is $1$ on $[s, t]$ and lies under two triangles of area $\delta/2$ each, so $$\int_0^1 f_\delta\,dx \le (t - s) + \delta.$$ So $\mu(K) \le t - s + \delta$ for every $\delta > 0$.
Together, $\mu([s, t]) = t - s$: the Riemann integral's measure gives every closed interval in $[0, 1]$ its length.
Replace $\int_0^1$ by $\int_{-\infty}^{\infty}$ (on $C_c(\mathbb{R})$ every integral is over a bounded interval anyway), and the Riesz Representation Theorem below hands you a measure on $\mathbb{R}$ that gives every interval its length. That's Lebesgue measure, and it's how Lebesgue Measure, the next page, builds it.
Point masses work the same way. For $\Lambda f = f(c)$: every $f \ge \chi_K$ has $f(c) = 1$ when $c \in K$, so $\mu(K) = 1$; when $c \notin K$, Urysohn's lemma with $V = X \setminus \{c\}$ gives $f$ with $f(c) = 0$, so $\mu(K) = 0$. Mix the two kinds and play with the result.
The one thing to watch is the endpoints. A spread of mass doesn't care whether an endpoint is in or out (a single point has length $0$). A point mass cares a lot. Now see how close your own functions can get. Squeeze from above onto a compact set, then fill an open set from below.
You can get as close as you like, but on the open set you never get past the mass inside it: the sup is approached, not reached. Your turn. Each functional below mixes point masses with a spread of mass, and you read off the measure of a set.
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We have a size for every set, and it's subadditive and finite on compact sets. Two lemmas are left before the theorem: additivity on $\mathcal{M}_F$, and the fact that $\mathcal{M}$ is a σ-algebra. The first one says that the sets in $\mathcal{M}_F$ behave the way you'd hope, and it gives us the squeeze between compact and open sets that the rest of this page is about.
(a) For every open $V$, $\mu(V) = \sup\{\mu(K) : K \subseteq V, \ K \text{ compact}\}$. So every open $V$ with $\mu(V) < \infty$ is in $\mathcal{M}_F$.
(b) If $E_1, E_2, \ldots \in \mathcal{M}_F$ are pairwise disjoint and $E = \bigcup E_i$, then $\mu(E) = \sum \mu(E_i)$. If also $\mu(E) < \infty$, then $E \in \mathcal{M}_F$.
(c) If $E \in \mathcal{M}_F$ and $\varepsilon > 0$, there are a compact $K$ and an open $V$ with $K \subseteq E \subseteq V$ and $\mu(V \setminus K) < \varepsilon$.
Proof. (a) Monotonicity gives $\le$. For $\ge$, let $\alpha < \mu(V)$ and pick $f$ under $V$ with $\Lambda f > \alpha$. Let $K = \operatorname{supp} f$, compact and inside $V$. If $W$ is any open set containing $K$, then $f$ is under $W$, so $\Lambda f \le \mu(W)$. Taking the inf over such $W$, $\Lambda f \le \mu(K)$, so $\mu(K) > \alpha$.
(b) Two disjoint compact sets. Let $K_1, K_2$ be compact and disjoint, and $\varepsilon > 0$. Urysohn's lemma with $K = K_1$ and $V = X \setminus K_2$ gives $f \in C_c(X)$ with $0 \le f \le 1$, $f = 1$ on $K_1$, and $f = 0$ on $K_2$. Lemma 2 gives $g \in C_c(X)$ with $\chi_{K_1 \cup K_2} \le g \le 1$ and $\Lambda g < \mu(K_1 \cup K_2) + \varepsilon$. Now $\chi_{K_1} \le fg \le 1$ and $\chi_{K_2} \le (1 - f)g \le 1$, so by Lemma 2 again, $$\mu(K_1) + \mu(K_2) \le \Lambda(fg) + \Lambda((1 - f)g) = \Lambda g < \mu(K_1 \cup K_2) + \varepsilon.$$ With Lemma 1, $\mu(K_1 \cup K_2) = \mu(K_1) + \mu(K_2)$, and by induction the same holds for finitely many disjoint compact sets.
The general case. Lemma 1 gives $\mu(E) \le \sum \mu(E_i)$. For $\ge$, let $\varepsilon > 0$. Since $E_i \in \mathcal{M}_F$, there is a compact $H_i \subseteq E_i$ with $\mu(H_i) > \mu(E_i) - \varepsilon/2^i$. The $H_i$ are disjoint, so with $K_n = H_1 \cup \cdots \cup H_n$, $$\mu(E) \ge \mu(K_n) = \sum_{i=1}^{n} \mu(H_i) > \sum_{i=1}^{n} \mu(E_i) - \varepsilon.$$ Let $n \to \infty$ and then $\varepsilon \to 0$: $\mu(E) \ge \sum \mu(E_i)$. If $\mu(E) < \infty$, choose $N$ with $\sum_{i \le N} \mu(E_i) > \mu(E) - \varepsilon$. Then the compact set $K_N \subseteq E$ has $\mu(K_N) > \mu(E) - 2\varepsilon$, so $E \in \mathcal{M}_F$.
(c) By the definitions of $\mu(E)$ and of $\mathcal{M}_F$, there are a compact $K \subseteq E$ and an open $V \supseteq E$ with $$\mu(V) - \tfrac{\varepsilon}{2} < \mu(E) < \mu(K) + \tfrac{\varepsilon}{2}.$$ The set $V \setminus K$ is open with $\mu(V \setminus K) \le \mu(V) < \infty$, so it's in $\mathcal{M}_F$ by (a). So is $K$ (Lemma 2). They are disjoint with union $V$, so by (b), $\mu(K) + \mu(V \setminus K) = \mu(V) < \mu(K) + \varepsilon$. Since $\mu(K)$ is finite, $\mu(V \setminus K) < \varepsilon$. $\blacksquare$
Part (c) is the squeeze: compact inside, open outside, almost nothing in between. It's the engine for the last lemma. A difference of two sets in $\mathcal{M}_F$ gets squeezed by squeezing both, and from there $\mathcal{M}$ becomes a σ-algebra.
(a) If $A, B \in \mathcal{M}_F$, then $A \setminus B$, $A \cup B$, and $A \cap B$ are in $\mathcal{M}_F$.
(b) $\mathcal{M}$ is a σ-algebra in $X$ that contains every Borel set.
(c) $\mathcal{M}_F = \{E \in \mathcal{M} : \mu(E) < \infty\}$.
(d) $\mu$, restricted to $\mathcal{M}$, is a positive measure.
Proof. (a) Let $\varepsilon > 0$. Lemma 3(c) gives compact $K_1, K_2$ and open $V_1, V_2$ with $K_1 \subseteq A \subseteq V_1$, $K_2 \subseteq B \subseteq V_2$, and $\mu(V_i \setminus K_i) < \varepsilon$. Then $$A \setminus B \subseteq V_1 \setminus K_2 \subseteq (V_1 \setminus K_1) \cup (K_1 \setminus V_2) \cup (V_2 \setminus K_2),$$ so by Lemma 1, $\mu(A \setminus B) < \varepsilon + \mu(K_1 \setminus V_2) + \varepsilon$. The set $K_1 \setminus V_2 = K_1 \cap V_2^c$ is compact (a closed subset of $K_1$) and lies in $A \setminus B$. So $A \setminus B$ has compact subsets of measure within $2\varepsilon$ of its own, and $\mu(A \setminus B) \le \mu(A) < \infty$. So $A \setminus B \in \mathcal{M}_F$. Then $A \cup B = (A \setminus B) \cup B$ is a disjoint union of two sets in $\mathcal{M}_F$ with finite measure, so it is in $\mathcal{M}_F$ by Lemma 3(b), and $A \cap B = A \setminus (A \setminus B)$.
(b) Let $K$ be compact; $K \in \mathcal{M}_F$ by Lemma 2. Whole space: $X \cap K = K$, so $X \in \mathcal{M}$. Complements: if $A \in \mathcal{M}$, then $A^c \cap K = K \setminus (A \cap K)$, a difference of two sets in $\mathcal{M}_F$, so it is in $\mathcal{M}_F$ by (a). Countable unions: if $A_1, A_2, \ldots \in \mathcal{M}$ and $A = \bigcup A_n$, let $B_1 = A_1 \cap K$ and $B_n = (A_n \cap K) \setminus (B_1 \cup \cdots \cup B_{n-1})$. By (a) each $B_n \in \mathcal{M}_F$, the $B_n$ are disjoint, and their union is $A \cap K$, which has measure at most $\mu(K) < \infty$. By Lemma 3(b), $A \cap K \in \mathcal{M}_F$. So $\mathcal{M}$ is a σ-algebra.
Borel sets: if $C$ is closed, $C \cap K$ is a closed subset of a compact set, so it is compact and in $\mathcal{M}_F$. So every closed set is in $\mathcal{M}$, and a σ-algebra containing the closed sets contains $\mathcal{B}(X)$.
(c) If $E \in \mathcal{M}_F$, then $E \cap K \in \mathcal{M}_F$ for every compact $K$ by (a), so $E \in \mathcal{M}$. Conversely, let $E \in \mathcal{M}$ with $\mu(E) < \infty$, and $\varepsilon > 0$. Choose an open $V \supseteq E$ with $\mu(V) < \infty$. Then $V \in \mathcal{M}_F$ by Lemma 3(a), so Lemma 3(c) gives a compact $K \subseteq V$ with $\mu(V \setminus K) < \varepsilon$. Since $E \cap K \in \mathcal{M}_F$, there is a compact $H \subseteq E \cap K$ with $\mu(E \cap K) < \mu(H) + \varepsilon$. As $E \subseteq (E \cap K) \cup (V \setminus K)$, $$\mu(E) \le \mu(E \cap K) + \mu(V \setminus K) < \mu(H) + 2\varepsilon.$$ So $E \in \mathcal{M}_F$.
(d) $\mu$ takes values in $[0, \infty]$ and $\mu(\emptyset) = 0$. Let $E_1, E_2, \ldots \in \mathcal{M}$ be pairwise disjoint. If some $\mu(E_i) = \infty$, then $\mu(\bigcup E_i) = \infty = \sum \mu(E_i)$ by monotonicity. Otherwise every $E_i \in \mathcal{M}_F$ by (c), and Lemma 3(b) gives $\mu(\bigcup E_i) = \sum \mu(E_i)$. $\blacksquare$
So $\mu$ is a measure on a σ-algebra that contains every Borel set. One thing left: does integrating against $\mu$ give back $\Lambda$? The idea is the Lebesgue sum. Cut the range of $f$ into thin levels, and let $E_i$ be the set where $f$ lands in the $i$-th level. Then $\int f\,d\mu \approx \sum y_i\,\mu(E_i)$, and a partition of unity splits $f$ into pieces that live near each $E_i$, so $\Lambda f$ is close to the same sum.
Finer levels make the error smaller, and in the limit $\Lambda f$ and $\int f\,d\mu$ are trapped together. Time to say it all at once.
$\Lambda$ doesn't know it's an integral yet. Nobody tell it.
Let $X$ be a locally compact Hausdorff space and $\Lambda$ a positive linear functional on $C_c(X)$. Then there is a σ-algebra $\mathcal{M}$ in $X$ that contains every Borel set, and a positive measure $\mu$ on $\mathcal{M}$, such that:
(a) $\Lambda f = \int_X f\,d\mu$ for every $f \in C_c(X)$;
(b) $\mu(K) < \infty$ for every compact $K \subseteq X$;
(c) for every $E \in \mathcal{M}$, $\mu(E) = \inf\{\mu(V) : E \subseteq V, \ V \text{ open}\}$;
(d) $\mu(E) = \sup\{\mu(K) : K \subseteq E, \ K \text{ compact}\}$ whenever $E$ is open, or $E \in \mathcal{M}$ with $\mu(E) < \infty$;
(e) if $E \in \mathcal{M}$, $A \subseteq E$, and $\mu(E) = 0$, then $A \in \mathcal{M}$.
Moreover, $\mu$ is the only positive measure on $\mathcal{M}$ with properties (a) through (d).
Proof. Take $\mu$, $\mathcal{M}_F$, and $\mathcal{M}$ as built above. By Lemma 4, $\mathcal{M}$ is a σ-algebra containing every Borel set and $\mu$ is a positive measure on it. Property (b) is Lemma 2, and (c) is the definition of $\mu(E)$. Property (d) is Lemma 3(a) for open sets and Lemma 4(c) for sets of finite measure. For (e): $\mu(A) \le \mu(E) = 0$, and every compact subset of $A$ has measure $0$, so $A \in \mathcal{M}_F \subseteq \mathcal{M}$ by Lemma 4(c).
(a). Every $f \in C_c(X)$ is $u + iv$ with $u, v$ real functions in $C_c(X)$, and both sides of (a) are linear, so it's enough to treat real $f$. It's also enough to prove $\Lambda f \le \int_X f\,d\mu$ for every real $f$: applied to $-f$, it gives the reverse inequality. (The integral is finite: $f$ is bounded and vanishes off $K = \operatorname{supp} f$, and $\mu(K) < \infty$.)
Let $f$ be real with support $K$, choose $a < b$ with $f(X) \subseteq [a, b]$, and let $\varepsilon > 0$. Pick numbers $y_0 < a < y_1 < \cdots < y_n = b$ with $y_i - y_{i-1} < \varepsilon$, and let $$E_i = \{x : y_{i-1} < f(x) \le y_i\} \cap K \qquad (i = 1, \ldots, n).$$ The $E_i$ are disjoint Borel sets (since $f$ is continuous) with union $K$, so $\sum \mu(E_i) = \mu(K)$. Choose open $V_i \supseteq E_i$ with $\mu(V_i) < \mu(E_i) + \varepsilon/n$; shrinking $V_i$ to $V_i \cap \{f < y_i + \varepsilon\}$, we may assume $f < y_i + \varepsilon$ on $V_i$. The partition of unity gives $h_i$ under $V_i$ with $\sum h_i = 1$ on $K$. Then $f = \sum h_i f$, and:
Since $|a| + y_i + \varepsilon \ge 0$, positivity of $\Lambda$ gives $$\begin{aligned} \Lambda f = \sum_{i=1}^{n} \Lambda(h_i f) &\le \sum_{i=1}^{n} (|a| + y_i + \varepsilon)\,\Lambda h_i - |a| \sum_{i=1}^{n} \Lambda h_i \\ &\le \sum_{i=1}^{n} (|a| + y_i + \varepsilon)\Big(\mu(E_i) + \frac{\varepsilon}{n}\Big) - |a|\,\mu(K) \\ &= \sum_{i=1}^{n} (y_i - \varepsilon)\,\mu(E_i) + 2\varepsilon\,\mu(K) + \frac{\varepsilon}{n} \sum_{i=1}^{n} (|a| + y_i + \varepsilon) \\ &\le \int_X f\,d\mu + \varepsilon\,\big(2\mu(K) + |a| + b + \varepsilon\big). \end{aligned}$$ Since $\varepsilon > 0$ was arbitrary, $\Lambda f \le \int_X f\,d\mu$.
Uniqueness. Let $\mu_1, \mu_2$ be positive measures on $\mathcal{M}$ with (a) through (d), $K$ compact, and $\varepsilon > 0$. By (b) and (c) there is an open $V \supseteq K$ with $\mu_2(V) < \mu_2(K) + \varepsilon$, and Urysohn's lemma gives $f$ with $\chi_K \le f \le \chi_V$. Then $$\mu_1(K) = \int \chi_K\,d\mu_1 \le \int f\,d\mu_1 = \Lambda f = \int f\,d\mu_2 \le \int \chi_V\,d\mu_2 = \mu_2(V) < \mu_2(K) + \varepsilon.$$ So $\mu_1(K) \le \mu_2(K)$, and by symmetry they are equal on compact sets. By (d) they then agree on open sets, and by (c) on all of $\mathcal{M}$. $\blacksquare$
Read it as a dictionary. Functionals on one side, measures on the other, and every positive functional has exactly one measure that's finite on compacts and regular enough. Notice the theorem says $\mathcal{M}$ contains the Borel sets. It can contain a lot more.
Let $\Lambda f = f(0) + f(1)$ on $C_c(\mathbb{R})$. Find the Riesz measure $\mu$ and the σ-algebra $\mathcal{M}$.
Open sets. If $f$ is under an open $V$, then $\Lambda f = f(0) + f(1)$ counts at most the points of $\{0, 1\}$ inside $V$ (each $f(x) \le 1$, and $f = 0$ off $V$), and Urysohn's lemma gives an $f$ under $V$ that is $1$ at each of them. So $\mu(V)$ is the number of points of $\{0, 1\}$ in $V$.
Every set. For any $E \subseteq \mathbb{R}$, the open set $V = \mathbb{R} \setminus (\{0, 1\} \setminus E)$ contains $E$ and holds exactly the points of $\{0, 1\}$ that $E$ holds, and no open $V \supseteq E$ holds fewer. So $\mu(E) = |E \cap \{0, 1\}|$, which is $(\delta_0 + \delta_1)(E)$.
The σ-algebra. Every $E$ has $\mu(E) \le 2 < \infty$, and the compact set $E \cap \{0, 1\} \subseteq E$ has the same measure as $E$. So every subset of $\mathbb{R}$ is in $\mathcal{M}_F$, and $\mathcal{M}$ is all of $\mathcal{P}(\mathbb{R})$.
So for point masses, every set is measurable, far past the Borel sets. Properties (c) and (d) are the fine print, and they're the subject of the rest of this page. Match each conclusion to the part of the theorem that gives it.
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Drop "positive" and allow any bounded linear functional on $C_0(X)$, and Riesz still works, with complex measures in place of positive ones. That's the version on Duality: (Lᵖ)* and C₀(X)*.
Riesz builds a measure on $\mathcal{M}$, but $\mathcal{M}$ depends on $\Lambda$. The σ-algebra every Riesz measure has in common is the Borel sets, so that's where we compare measures.
Let $X$ be a locally compact Hausdorff space. A Borel measure on $X$ is a positive measure defined on the Borel σ-algebra $\mathcal{B}(X)$.
Every Riesz measure, cut down to $\mathcal{B}(X)$, is a Borel measure. What made the proof work was approximating sets: open sets from outside, compact sets from inside. Those two kinds of approximation get names.
Let $\mu$ be a positive measure on a σ-algebra in $X$ that contains $\mathcal{B}(X)$. A set $E$ in that σ-algebra is outer regular if $\mu(E) = \inf\{\mu(V) : E \subseteq V, \ V \text{ open}\}$.
Outer regular means you can cover $E$ by open sets that are barely bigger in measure. That can fail, even for a single point.
On $\mathbb{R}$, let $\mu = \sum_{n=1}^{\infty} \delta_{1/n}$ (one unit of mass at each point $1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots$). Show that $\{0\}$ is not outer regular.
$\mu(\{0\}) = 0$, since $0$ is not one of the points $1/n$. But every open $V \ni 0$ contains an interval $(-r, r)$, and that holds every $1/n$ with $n > 1/r$: infinitely many units of mass. So $\mu(V) = \infty$ for every open $V \supseteq \{0\}$, and $\inf \mu(V) = \infty \neq 0$.
The culprit is mass piling up at a point: the compact set $[0, 1]$ has infinite measure. Hold on to that, it comes back at the end of the page. The inside version of regularity uses compact sets.
Let $\mu$ be a positive measure on a σ-algebra in $X$ that contains $\mathcal{B}(X)$. A set $E$ in that σ-algebra is inner regular if $\mu(E) = \sup\{\mu(K) : K \subseteq E, \ K \text{ compact}\}$.
Here's the squeeze, on a set with both. Compact sets grow inside it, open sets shrink around it, and the measures meet in the middle.
A measure where every Borel set can be squeezed like that is the best kind to have around.
A positive measure $\mu$ on a σ-algebra in $X$ that contains $\mathcal{B}(X)$ is regular if every Borel set is both outer regular and inner regular.
The Riesz measure is outer regular on all of $\mathcal{M}$, but property (d) only promises inner regularity for open sets and sets of finite measure. A single point mass passes every test.
Show that $\delta_0$ is a regular measure on $\mathbb{R}$.
Let $E$ be a Borel set. If $0 \in E$, then $\delta_0(E) = 1$. Every open $V \supseteq E$ also has $\delta_0(V) = 1$, and the compact set $\{0\} \subseteq E$ has $\delta_0(\{0\}) = 1$. If $0 \notin E$, then $\delta_0(E) = 0$, the open set $V = \mathbb{R} \setminus \{0\}$ contains $E$ with $\delta_0(V) = 0$, and the compact set $\emptyset$ gives the sup $0$. Either way, $E$ is outer and inner regular.
Finitely many point masses work the same way. Now go break outer regularity yourself.
Counting measure on $\mathbb{R}$: $\mu(\{0\}) = 1$, and every open set around $0$ costs $\infty$. Outer regularity, but make it a scam.
Every failure you found had the same shape: a set whose open neighborhoods all grab infinitely much more mass than the set itself.
Don't read more into the theorem than it says. Riesz gives outer regularity for every set in $\mathcal{M}$, but inner regularity only for open sets and sets of finite measure. That gap is real: there are locally compact Hausdorff spaces where a Riesz measure has a Borel set of infinite measure whose compact subsets all have measure $0$ (Rudin, Real and Complex Analysis, Exercise 2.17). The next section shows the gap closes as soon as $X$ is a countable union of compact sets.
Infinite measure is where inner regularity can go wrong, so the fix is to chop $X$ into countably many pieces of finite measure. Compact pieces are the natural choice, since Riesz makes them finite.
A set $E$ in a topological space is σ-compact if $E = \bigcup_{n=1}^{\infty} K_n$ for some compact sets $K_1, K_2, \ldots$.
Most spaces you know pass. One natural space doesn't.
Show that (a) $\mathbb{R}^n$ is σ-compact, (b) every open $U \subseteq \mathbb{R}^n$ is σ-compact, and (c) an uncountable set $X$ with the discrete topology is locally compact Hausdorff but not σ-compact.
(a) $\mathbb{R}^n = \bigcup_{j} \overline{B}(0, j)$, and closed balls are compact (Heine–Borel).
(b) Take the closed balls $\overline{B}(q, r) \subseteq U$ with $q \in \mathbb{Q}^n$ and $r$ rational. There are countably many. Each $x \in U$ has some $B(x, 2r) \subseteq U$ with $r$ rational; pick $q \in \mathbb{Q}^n$ with $|x - q| < r$. Then $x \in \overline{B}(q, r) \subseteq B(x, 2r) \subseteq U$. So these balls cover $U$.
(c) Each $\{x\}$ is open with compact closure $\{x\}$, and distinct points have disjoint open neighborhoods. But a compact set is covered by its own singletons, so it is finite, and a countable union of finite sets is countable.
When $X = \bigcup K_n$ and $\mu(K_n) < \infty$, the whole space is a countable union of finite-measure pieces, which has its own name.
A measurable set $E$ in a measure space has σ-finite measure if $E = \bigcup_{n=1}^{\infty} E_n$ for some measurable sets $E_n$ with $\mu(E_n) < \infty$. If $X$ itself has σ-finite measure, $\mu$ is called σ-finite.
So on a σ-compact space, every Riesz measure is σ-finite. With that, the squeeze gets an upgrade. Compact and open sets can sandwich every set in $\mathcal{M}$, and if we take countably many slices of bread, the gap shrinks to nothing. The bread has names.
A set in a topological space is an $F_\sigma$ set if it is a countable union of closed sets.
The letters are French and German: fermé (closed) and Summe (union). The other one swaps both words.
A set in a topological space is a $G_\delta$ set if it is a countable intersection of open sets.
That's Gebiet (open region) and Durchschnitt (intersection). The complement of an $F_\sigma$ is a $G_\delta$ and vice versa, and every closed set is an $F_\sigma$ (a union of one set). Here are a few of each.
(a) Write $(0, 1]$ as a countable union of closed sets and as a countable intersection of open sets. (b) Show that $\mathbb{Q}$ is an $F_\sigma$ and $\mathbb{R} \setminus \mathbb{Q}$ is a $G_\delta$.
(a) $(0, 1] = \bigcup_{n=1}^{\infty} [\tfrac{1}{n}, 1]$, a union of closed intervals, and $(0, 1] = \bigcap_{n=1}^{\infty} (0, 1 + \tfrac{1}{n})$, an intersection of open intervals. So $(0, 1]$ is both, though it is neither open nor closed.
(b) $\mathbb{Q}$ is countable, and each $\{q\}$ is closed, so $\mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \{q\}$ is an $F_\sigma$. Its complement is $\bigcap_{q \in \mathbb{Q}} (\mathbb{R} \setminus \{q\})$, a countable intersection of open sets.
Half-open intervals manage both. Is every closed set also a $G_\delta$? In a metric space, yes.
Let $X$ be a metric space. Prove that every closed set in $X$ is a $G_\delta$, and that every open set in $X$ is an $F_\sigma$.
Proof. The empty set is open, so it is a $G_\delta$. Let $C \neq \emptyset$ be closed. The distance $d(x, C) = \inf\{d(x, y) : y \in C\}$ is continuous in $x$ (it changes by at most $d(x, x')$ when $x$ moves to $x'$; Question 2.3 on the Urysohn page), so each $V_n = \{x : d(x, C) < \tfrac{1}{n}\}$ is open. Then $$\bigcap_{n=1}^{\infty} V_n = \{x : d(x, C) = 0\} = \overline{C} = C.$$ So $C$ is a $G_\delta$. If $U$ is open, $U^c$ is closed, so $U^c = \bigcap V_n$ with $V_n$ open, and $U = \bigcup V_n^c$ is a countable union of closed sets. $\blacksquare$
So in $\mathbb{R}$, open sets, closed sets, half-open intervals, and countable sets are all $F_\sigma$, and the first three are $G_\delta$ too. That makes it tempting to think every nice set is both. It isn't, and the standard counterexample is the rationals.
Let $D \subseteq \mathbb{R}$ be countable and dense in some open interval $I$ (every open subinterval of $I$ meets $D$). Then $D$ is an $F_\sigma$ but not a $G_\delta$, and $\mathbb{R} \setminus D$ is a $G_\delta$ but not an $F_\sigma$. In particular, $\mathbb{Q}$ is an $F_\sigma$ and not a $G_\delta$.
Proof. List $D = \{d_1, d_2, \ldots\}$. Each $\{d_n\}$ is closed, so $D = \bigcup \{d_n\}$ is an $F_\sigma$, and $\mathbb{R} \setminus D$ is a $G_\delta$.
Suppose $D = \bigcap_{n=1}^{\infty} V_n$ with each $V_n$ open. Each $V_n$ contains $D$, so it meets every open subinterval of $I$. Build closed intervals $J_1 \supseteq J_2 \supseteq \cdots$ of positive length:
The $J_n$ are nested, nonempty, and compact, so they have a common point $x$ (the finite intersection property). Then $x \in J_n \subseteq V_n$ for every $n$, so $x \in D$, say $x = d_k$. But $d_k \notin J_k$, a contradiction. So $D$ is not a $G_\delta$.
If $\mathbb{R} \setminus D$ were an $F_\sigma$, its complement $D$ would be a $G_\delta$. $\blacksquare$
The proof is a game. Someone claims $D = \bigcap V_n$, and you keep choosing smaller and smaller intervals, each one inside the next $V_n$ and dodging the next point of $D$. Whatever is left at the end is in every $V_n$ and is none of the $d_n$. Watch it on $\mathbb{Q}$, then play it.
Your turn to be the dodger. The list never runs out, but neither do your intervals: each one is a closed interval of positive length.
The exercise on Locally Compact Spaces & Urysohn's Lemma showed that the points where a function $f\colon \mathbb{R} \to \mathbb{C}$ is continuous always form a $G_\delta$ (the set where the oscillation is $0$). And $\mathbb{Q}$ is not a $G_\delta$. So no function on $\mathbb{R}$ is continuous at exactly the rationals. (Exactly the irrationals is fine: try $f(p/q) = 1/q$ and $f = 0$ off $\mathbb{Q}$.)
$\mathbb{Q}$ is an $F_\sigma$ because it's a countable pile of points. It's not a $G_\delta$ because every open set it tries brings irrationals along.
So $F_\sigma$ and $G_\delta$ are genuinely different, and a set can be one without the other. Now the theorem. On a σ-compact space, every set in $\mathcal{M}$ sits between a closed set and an open set with almost nothing between them, and then between an $F_\sigma$ and a $G_\delta$ with nothing at all between them.
Let $X$ be a locally compact, σ-compact Hausdorff space, and let $\Lambda$, $\mathcal{M}$, and $\mu$ be as in the Riesz Representation Theorem. Then:
(a) for every $E \in \mathcal{M}$ and $\varepsilon > 0$, there are a closed $F$ and an open $V$ with $F \subseteq E \subseteq V$ and $\mu(V \setminus F) < \varepsilon$;
(b) $\mu$ is regular;
(c) for every $E \in \mathcal{M}$, there are an $F_\sigma$ set $A$ and a $G_\delta$ set $B$ with $A \subseteq E \subseteq B$ and $\mu(B \setminus A) = 0$.
Proof. (a) Write $X = \bigcup_{n=1}^{\infty} K_n$ with each $K_n$ compact. Let $E \in \mathcal{M}$ and $\varepsilon > 0$. Each $K_n \cap E$ is in $\mathcal{M}$ and has $\mu(K_n \cap E) \le \mu(K_n) < \infty$. By outer regularity there is an open $V_n \supseteq K_n \cap E$ with $$\mu\big(V_n \setminus (K_n \cap E)\big) = \mu(V_n) - \mu(K_n \cap E) < \frac{\varepsilon}{2^{n+1}}.$$ Let $V = \bigcup V_n$, which is open. Since $E = \bigcup (K_n \cap E)$, we have $V \setminus E \subseteq \bigcup \big(V_n \setminus (K_n \cap E)\big)$, so $\mu(V \setminus E) < \sum \varepsilon/2^{n+1} = \varepsilon/2$.
Apply the same argument to $E^c \in \mathcal{M}$: there is an open $W \supseteq E^c$ with $\mu(W \setminus E^c) < \varepsilon/2$. Let $F = W^c$, which is closed and lies in $E$. Then $E \setminus F = E \cap W = W \setminus E^c$, so $\mu(E \setminus F) < \varepsilon/2$, and $$\mu(V \setminus F) \le \mu(V \setminus E) + \mu(E \setminus F) < \varepsilon.$$
(b) Every Borel set is outer regular by the Riesz theorem (c). Inner regularity is the exercise below.
(c) By (a), for each $n$ there are a closed $F_n$ and an open $V_n$ with $F_n \subseteq E \subseteq V_n$ and $\mu(V_n \setminus F_n) < \tfrac{1}{n}$. Let $A = \bigcup F_n$, an $F_\sigma$, and $B = \bigcap V_n$, a $G_\delta$. Then $A \subseteq E \subseteq B$, and $B \setminus A \subseteq V_n \setminus F_n$ for every $n$, so $\mu(B \setminus A) < \tfrac{1}{n}$ for every $n$. Hence $\mu(B \setminus A) = 0$. $\blacksquare$
The proof of (a) spends the error budget the way you'd pay off a debt in installments: half of $\varepsilon$ split over the pieces $K_n$, a quarter, an eighth, and so on, so the total stays under $\varepsilon/2$.
Part (b) is yours. Part (a) already gives a closed set inside $E$ that's almost all of $E$, and closed sets in a σ-compact space are countable unions of compact ones.
Prove part (b) of the theorem: under its hypotheses, every Borel set $E$ is inner regular.
Proof. Let $\varepsilon > 0$. By (a) there is a closed $F \subseteq E$ with $\mu(E \setminus F) < \varepsilon$. Let $F_n = F \cap (K_1 \cup \cdots \cup K_n)$. Each $F_n$ is a closed subset of a compact set, so it is compact, $F_n \subseteq E$, and $F_n$ increases to $F$. By continuity from below, $\mu(F_n) \to \mu(F)$.
If $\mu(E) < \infty$, then $\mu(F) > \mu(E) - \varepsilon$, so some $\mu(F_n) > \mu(E) - \varepsilon$. If $\mu(E) = \infty$, then $\mu(F) = \infty$ (otherwise $\mu(E) \le \mu(F) + \varepsilon$ would be finite), so $\mu(F_n) \to \infty$. Either way, $\sup\{\mu(K) : K \subseteq E \text{ compact}\} = \mu(E)$. $\blacksquare$
Part (c) is the sandwich from Measures: there, the completion's sets were the ones caught between $A \subseteq E \subseteq B$ with $\mu(B \setminus A) = 0$. Here the bread is always Borel, so on a σ-compact space every set in $\mathcal{M}$ is an $F_\sigma$ plus a subset of a Borel null set, and $\mathcal{M}$ sits inside the completion of $\mu$ on $\mathcal{B}(X)$.
In $\mathbb{R}$ you can see (a) happen for the irrationals in $[0, 1]$, the $G_\delta$ that isn't an $F_\sigma$. They contain no interval, yet a closed subset of them can have length as close to $1$ as you like. (This uses the length measure from the example above; the next page makes it official.)
The closed set you built is full of gaps, but its gaps add up to almost nothing. Decide where each set sits among the four possibilities.
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So far regularity came from the construction: the Riesz measure is regular because we built it out of open and compact sets. What about a Borel measure that someone hands you, built any old way? If every open set is σ-compact (as in $\mathbb{R}^n$), being finite on compact sets is enough. The key step compares the given measure with the Riesz measure of its own integral, on open sets first.
Let $X$ be a locally compact Hausdorff space, and let $\lambda$ and $\mu$ be positive measures on σ-algebras containing $\mathcal{B}(X)$, both finite on compact sets, with $\int_X f\,d\lambda = \int_X f\,d\mu$ for every $f \in C_c(X)$. Prove that $\lambda(V) = \mu(V)$ for every σ-compact open set $V$.
Proof. Write $V = \bigcup_{i=1}^{\infty} K_i$ with each $K_i$ compact. Urysohn's lemma gives $f_i \in C_c(X)$ with $\chi_{K_i} \le f_i \le \chi_V$ and $\operatorname{supp} f_i \subseteq V$. Let $g_n = \max\{f_1, \ldots, f_n\}$. Then $g_n$ is continuous, its support lies in the compact set $\operatorname{supp} f_1 \cup \cdots \cup \operatorname{supp} f_n$, so $g_n \in C_c(X)$, and $0 \le g_1 \le g_2 \le \cdots$.
Also $g_n \to \chi_V$ pointwise: if $x \notin V$, every $f_i(x) = 0$; if $x \in V$, then $x \in K_i$ for some $i$, and $g_n(x) = 1$ for all $n \ge i$. By the Monotone Convergence Theorem, applied to each measure, $$\lambda(V) = \lim_{n \to \infty} \int_X g_n\,d\lambda = \lim_{n \to \infty} \int_X g_n\,d\mu = \mu(V). \qquad \blacksquare$$
Here's that limit: Urysohn functions for bigger and bigger compact pieces of $V$, stacked with a max, climbing to $\chi_V$. Both integrals ride along.
Agreeing on open sets is half the battle. Part (a) of the last theorem squeezes every Borel set between a closed set and an open set, and the difference of those is open, so the agreement spreads to everything.
Let $X$ be a locally compact Hausdorff space in which every open set is σ-compact. If $\lambda$ is a positive Borel measure on $X$ with $\lambda(K) < \infty$ for every compact $K$, then $\lambda$ is regular.
Proof. Since $\lambda$ is finite on compact sets, $C_c(X) \subseteq L^1(\lambda)$ and $\Lambda f = \int_X f\,d\lambda$ is a positive linear functional on $C_c(X)$ (the corollary on the Urysohn page). Let $\mu$ on $\mathcal{M} \supseteq \mathcal{B}(X)$ be its Riesz measure, so $\int f\,d\lambda = \int f\,d\mu$ for every $f \in C_c(X)$, and $\mu$ is finite on compact sets. The space $X$ is itself open, so it is σ-compact, and the last theorem applies to $\mu$: $\mu$ is regular. By the exercise above, $\lambda(V) = \mu(V)$ for every open $V$.
Let $E$ be a Borel set and $\varepsilon > 0$. Part (a) of the last theorem gives a closed $F$ and an open $V$ with $F \subseteq E \subseteq V$ and $\mu(V \setminus F) < \varepsilon$. The set $V \setminus F$ is open, so $\lambda(V \setminus F) = \mu(V \setminus F) < \varepsilon$ too. Hence $$\lambda(E) \le \lambda(V) = \mu(V) \le \mu(E) + \mu(V \setminus E) \le \mu(E) + \varepsilon,$$ $$\mu(E) \le \mu(V) = \lambda(V) \le \lambda(E) + \lambda(V \setminus E) \le \lambda(E) + \varepsilon.$$ If one of $\lambda(E)$, $\mu(E)$ is infinite, these force the other to be infinite too; otherwise $|\lambda(E) - \mu(E)| \le \varepsilon$ for every $\varepsilon > 0$. Either way $\lambda(E) = \mu(E)$. So $\lambda = \mu$ on $\mathcal{B}(X)$, and $\lambda$ is regular because $\mu$ is. $\blacksquare$
That turns regularity into a single check. Try it on a pile-up that looks dangerous.
Is $\mu = \sum_{n=1}^{\infty} 2^{-n}\,\delta_{1/n}$ a regular Borel measure on $\mathbb{R}$? Check $\{0\}$ directly too.
Every compact $K \subseteq \mathbb{R}$ has $\mu(K) \le \mu(\mathbb{R}) = \sum 2^{-n} = 1 < \infty$, and every open subset of $\mathbb{R}$ is σ-compact, so the theorem says $\mu$ is regular.
Directly: $\mu(\{0\}) = 0$, and the open set $(-r, r)$ holds only the points $1/n$ with $n > 1/r$, so $\mu((-r, r)) \le \sum_{n > 1/r} 2^{-n} \le 2^{1 - 1/r}$, which goes to $0$ as $r \to 0$. So $\{0\}$ is outer regular, unlike for $\sum \delta_{1/n}$.
So on $\mathbb{R}^n$, there's no such thing as an irregular Borel measure that's finite on bounded sets. The hypothesis "finite on compact sets" is the one doing the work, and it can't be dropped: $\sum \delta_{1/n}$ from the example above is a Borel measure on $\mathbb{R}$, every open subset of $\mathbb{R}$ is σ-compact, and it still fails outer regularity at $\{0\}$, because $\mu([0, 1]) = \infty$. With that one test in hand, sort these.
Infinite total mass is fine (counting measure on $\mathbb{Z}$ is regular). Mass piling up near a point is fine too, as long as the pile is finite. What breaks regularity is an infinite pile inside a compact set. Spot it.
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Last in the lesson: the proofs. Every one on this page has a step where the right move is one small choice (inf or sup, union or intersection, $\varepsilon/2^n$ or $\varepsilon$). Pick it.
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Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.
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Next, the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; most types here need their answer choices, and they'll say so.
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Last, the Spicy Bonus. A regular probability measure on a compact space has a smallest compact set that carries all of its mass. Finding it uses compactness twice and outer regularity once.
Let $X$ be a compact Hausdorff space and $\mu$ a regular Borel measure on $X$ with $\mu(X) = 1$. Prove that there is a compact set $K \subseteq X$ (the support of $\mu$) such that $\mu(K) = 1$ but $\mu(H) < 1$ for every compact $H \subsetneq K$. Show also that $X \setminus K$ is the largest open set of measure $0$: it has measure $0$ and contains every open set of measure $0$.
Hint: let $K$ be the intersection of all compact sets of measure $1$, and show every open set containing $K$ has measure $1$.
Proof. Let $\{K_\alpha\}$ be the family of all compact sets with $\mu(K_\alpha) = 1$; it is nonempty since $X$ is one. Let $K = \bigcap_\alpha K_\alpha$. Compact sets in a Hausdorff space are closed, so $K$ is closed in the compact space $X$, hence compact.
$\mu(K) = 1$. Let $V$ be open with $K \subseteq V$. Then $V^c$ is closed in $X$, so compact, and $V^c \subseteq K^c = \bigcup_\alpha K_\alpha^c$, an open cover. Finitely many cover it: $V^c \subseteq K_{\alpha_1}^c \cup \cdots \cup K_{\alpha_n}^c$. Each $\mu(K_{\alpha_i}^c) = 1 - 1 = 0$, so $\mu(V^c) = 0$ and $\mu(V) = 1$. Since $K$ is compact, hence Borel, it is outer regular: $\mu(K) = \inf\{\mu(V) : K \subseteq V \text{ open}\} = 1$.
Proper compact subsets. If $H \subsetneq K$ is compact with $\mu(H) = 1$, then $H$ is one of the $K_\alpha$, so $K \subseteq H$, contradicting $H \neq K$. So $\mu(H) < 1$.
The largest open null set. $\mu(X \setminus K) = 1 - \mu(K) = 0$. If $W$ is open with $\mu(W) = 0$, then $W^c$ is compact with $\mu(W^c) = 1$, so $W^c$ is one of the $K_\alpha$ and $K \subseteq W^c$. That is, $W \subseteq X \setminus K$. $\blacksquare$
(Only outer regularity of compact sets was used. The same $K$ works for any Borel probability measure on a compact metric space, where every Borel measure that's finite on compact sets is regular by the last theorem, since open sets of a compact metric space are σ-compact.)