Every page so far has quietly assumed there's a measure on $\mathbb{R}$ that gives an interval its length. Time to pay for that. This page builds it, in every dimension at once: a measure $m$ on $\mathbb{R}^n$ that gives a box its volume. Then it checks the two things a "volume" ought to do, which are not caring where a set sits and scaling correctly when you stretch it. It ends with the bad news: some sets have no volume at all.
On The Lebesgue Integral we borrowed one fact, that some measure $m$ on the Borel sets has $m([a, b]) = b - a$, and promised to construct it later. This is later.
Here's the wish list. We want a measure on $\mathbb{R}^n$ that agrees with the volume you already know on boxes, and that doesn't change when you slide a set around. Sliding needs a name first.
Let $E \subseteq \mathbb{R}^n$ and $x \in \mathbb{R}^n$. The translate of $E$ by $x$ is the set $$E + x = \{y + x : y \in E\}.$$
Same shape, new address.
Find $E + x$ for $E = [0, 1) \cup \{5\}$ and $x = -3$, and for the unit disk $D = \{y \in \mathbb{R}^2 : |y| < 1\}$ and $x = (2, -1)$.
Subtract $3$ from every point: $E - 3 = [-3, -2) \cup \{2\}$.
$D + (2, -1) = \{y + (2, -1) : |y| < 1\} = \{z : |z - (2, -1)| < 1\}$, the open disk of radius $1$ centered at $(2, -1)$.
Next, the sets whose volume nobody argues about. In one dimension they're intervals, in two they're rectangles, and in general they're products of intervals.
An $n$-cell in $\mathbb{R}^n$ is a set of the form $W = I_1 \times I_2 \times \cdots \times I_n$, where each $I_i$ is a bounded interval with endpoints $\alpha_i \le \beta_i$.
Each $I_i$ can be open, closed, or half-open, and it can even be a single point ($\alpha_i = \beta_i$). The ends won't matter to the volume, and that's the whole point of the next definition.
The volume of the $n$-cell $W = I_1 \times \cdots \times I_n$, where $I_i$ has endpoints $\alpha_i \le \beta_i$, is $$\operatorname{vol}(W) = \prod_{i=1}^{n} (\beta_i - \alpha_i).$$
So volume is just "multiply the side lengths", whatever the ends look like.
Find the volumes of $W_1 = [0, 2) \times (1, 4] \times [-1, 3]$ in $\mathbb{R}^3$ and of $W_2 = [0, 3] \times [2, 2]$ in $\mathbb{R}^2$.
$W_1$ has sides of length $2 - 0 = 2$, $4 - 1 = 3$, and $3 - (-1) = 4$, so $\operatorname{vol}(W_1) = 2 \cdot 3 \cdot 4 = 24$. The open and closed ends make no difference.
$W_2$ is a segment: its second side $[2, 2]$ is a single point, of length $0$. So $\operatorname{vol}(W_2) = 3 \cdot 0 = 0$.
A cell with one flat side, like $W_2$, has volume $0$. We'll lean on that later. One more kind of cell: the ones that tile space without overlapping. Half-open intervals are what make that work, since $[0, 1)$ and $[1, 2)$ meet nowhere.
Let $\alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbb{R}^n$ and $\delta > 0$. The $\delta$-box with corner at $\alpha$ is $$Q(\alpha, \delta) = [\alpha_1, \alpha_1 + \delta) \times \cdots \times [\alpha_n, \alpha_n + \delta).$$
The useful ones sit on a grid. Chop space into boxes of side $2^{-j}$ with corners at multiples of $2^{-j}$, and you get a tiling that refines itself every time $j$ goes up by one.
A dyadic box of level $j$ (for $j \in \mathbb{N}$) is a $2^{-j}$-box whose corner has every coordinate a multiple of $2^{-j}$: $Q(2^{-j} a, 2^{-j})$ with $a \in \mathbb{Z}^n$.
The level-$j$ boxes cover $\mathbb{R}^n$ without overlapping (every coordinate of a point lies in exactly one $[a 2^{-j}, (a + 1) 2^{-j})$). And two dyadic boxes of different levels never partly overlap: the smaller one is either inside the bigger one or misses it entirely.
Write the open interval $(0, 1)$ as a disjoint union of dyadic boxes.
At level $1$ the boxes are $[0, \tfrac{1}{2})$ and $[\tfrac{1}{2}, 1)$. The second fits inside $(0, 1)$; the first doesn't, because it contains $0$. At level $2$, the boxes not already covered are $[0, \tfrac{1}{4})$ and $[\tfrac{1}{4}, \tfrac{1}{2})$, and only $[\tfrac{1}{4}, \tfrac{1}{2})$ fits. The pattern repeats: $$(0, 1) = [\tfrac{1}{2}, 1) \cup [\tfrac{1}{4}, \tfrac{1}{2}) \cup [\tfrac{1}{8}, \tfrac{1}{4}) \cup \cdots = \bigcup_{j=1}^{\infty} [2^{-j}, 2^{-j+1}).$$ Every $x \in (0, 1)$ lands in exactly one piece, and no piece contains $0$.
That greedy recipe (take every box that fits and isn't already covered, one level at a time) works for every open set in every dimension. This lemma is the workhorse of the page; the notes use it without stating it.
Every open set $V \subseteq \mathbb{R}^n$ is a countable union of pairwise disjoint dyadic boxes.
Proof. Two facts about dyadic boxes first.
(i) If $Q$ has level $j$ and $Q'$ has level $i \ge j$, then $Q' \subseteq Q$ or $Q \cap Q' = \emptyset$. Indeed, each side of $Q'$ is $[a 2^{-i}, (a + 1) 2^{-i})$ and each side of $Q$ is $[b 2^{-j}, (b + 1) 2^{-j})$. Since $2^{-j}$ is a whole multiple of $2^{-i}$, the endpoints of the bigger interval are on the smaller one's grid, so the smaller interval is inside the bigger one or misses it. A box is inside another exactly when every side is.
(ii) If $x \in V$, then for all large $j$ the level-$j$ box containing $x$ lies inside $V$. Indeed, $V$ is open, so some ball $B(x, r)$ lies in $V$. The level-$j$ box containing $x$ has diameter $\sqrt{n}\, 2^{-j}$, so once $\sqrt{n}\, 2^{-j} < r$, every point of that box is within $r$ of $x$, and the box lies in $B(x, r) \subseteq V$.
Now let $\mathcal{F}_1$ be the level-$1$ boxes inside $V$, and for $j \ge 2$ let $\mathcal{F}_j$ be the level-$j$ boxes inside $V$ that are not inside any box of $\mathcal{F}_1 \cup \cdots \cup \mathcal{F}_{j-1}$. There are countably many boxes of each level, so $\mathcal{F} = \bigcup_j \mathcal{F}_j$ is countable, and every box in it lies in $V$.
They cover $V$. Let $x \in V$. By (ii), some level-$j$ box $Q \ni x$ lies in $V$. Either $Q \in \mathcal{F}_j$, or $Q$ lies inside an earlier chosen box. Either way $x$ is covered.
They are disjoint. Let $Q \in \mathcal{F}_j$ and $Q' \in \mathcal{F}_i$ be different, with $i \ge j$. If $i = j$, different boxes of one level are disjoint. If $i > j$, then by (i) either $Q' \subseteq Q$, which the definition of $\mathcal{F}_i$ rules out, or $Q \cap Q' = \emptyset$. $\blacksquare$
Here's the recipe running on a blob in the plane. Each level only adds the boxes that fit and aren't already inside a bigger chosen box.
The covered area climbs toward the blob's area, and nothing is ever counted twice. That second part is the reason for the lemma: a measure adds up over disjoint pieces, so if we know a measure on boxes, we know it on every open set. Try the recipe by hand.
On σ-Algebras & Measurable Functions, every open set in $\mathbb{R}^2$ was a countable union of rational rectangles. That was enough to make open sets measurable, but those rectangles overlap, so their areas don't add up to anything useful. Dyadic boxes fix exactly that.
Big boxes first is the efficient way: a box you add early covers ground that would otherwise take dozens of small ones. And no matter how fine you go, the boxes never quite reach the edge of the blob. They don't have to, since countably many of them get there in the limit.
From The Riesz Representation Theorem: a positive linear functional $\Lambda$ on $C_c(X)$, for $X$ locally compact Hausdorff, comes from a complete measure $\mu$ on a σ-algebra containing the Borel sets, with $\Lambda f = \int f\,d\mu$ and $\mu(K) < \infty$ for compact $K$. When $X$ is also σ-compact, $\mu$ is regular, and every measurable set lies between an $F_\sigma$ and a $G_\delta$ that differ by a null set. And when every open set of $X$ is σ-compact, every Borel measure that is finite on compact sets is regular.
The Riesz Representation Theorem turns a functional into a measure, so all we need is the right functional. For continuous functions with compact support, there's an integral everyone already trusts: the Riemann integral. Feed it to Riesz, and out comes Lebesgue measure. Riesz hands us three things, which we'll use by name: $m(K) < \infty$ for compact $K$; regularity; and the sandwich between an $F_\sigma$ (a countable union of closed sets) and a $G_\delta$ (a countable intersection of open sets). The work is checking that the measure gives cells their volume.
There is a σ-algebra $\mathcal{M}$ in $\mathbb{R}^n$ containing every Borel set, and a complete positive measure $m$ on $\mathcal{M}$, such that:
(a) $m(W) = \operatorname{vol}(W)$ for every $n$-cell $W$;
(b) $E \in \mathcal{M}$ if and only if there are an $F_\sigma$ set $A$ and a $G_\delta$ set $B$ with $A \subseteq E \subseteq B$ and $m(B \setminus A) = 0$. Moreover, $m$ is regular.
Step 1: the measure. For $f \in C_c(\mathbb{R}^n)$, let $\Lambda f$ be the Riemann integral of $f$ over any cell containing $\operatorname{supp} f$. (A continuous function on a closed cell is Riemann integrable, and the value doesn't depend on which cell, since $f = 0$ outside its support.) $\Lambda$ is linear, and $\Lambda f \ge 0$ when $f \ge 0$. $\mathbb{R}^n$ is locally compact Hausdorff, so the Riesz Representation Theorem gives a σ-algebra $\mathcal{M}$ containing the Borel sets and a complete positive measure $m$ on it with $\Lambda f = \int f\,dm$ for every $f \in C_c(\mathbb{R}^n)$ and $m(K) < \infty$ for every compact $K$.
Step 2: open cells. Let $W = (\alpha_1, \beta_1) \times \cdots \times (\alpha_n, \beta_n)$ with every $\alpha_i < \beta_i$. For large $j$, let $E_j = [\alpha_1 + \tfrac{1}{j}, \beta_1 - \tfrac{1}{j}] \times \cdots \times [\alpha_n + \tfrac{1}{j}, \beta_n - \tfrac{1}{j}]$, a closed cell inside $W$. Urysohn's lemma gives $f_j \in C_c(\mathbb{R}^n)$ with $\chi_{E_j} \le f_j \le \chi_W$. Since $\chi_{E_j} \le f_j \le \chi_W$ pointwise, monotonicity of the Riemann integral gives $$\operatorname{vol}(E_j) \le \Lambda f_j \le \operatorname{vol}(W).$$ As $j \to \infty$, $\operatorname{vol}(E_j) \to \operatorname{vol}(W)$, so $\Lambda f_j \to \operatorname{vol}(W)$. On the other side, $f_j(x) \to \chi_W(x)$ for every $x$ (a point of $W$ is in $E_j$ for all large $j$, and a point outside $W$ has $f_j = 0$), and $0 \le f_j \le \chi_{\overline{W}}$ with $m(\overline{W}) < \infty$. By dominated convergence, $$m(W) = \int \chi_W\,dm = \lim_{j} \int f_j\,dm = \lim_j \Lambda f_j = \operatorname{vol}(W).$$
Step 3: every cell. Let $W$ be any $n$-cell with sides from $\alpha_i$ to $\beta_i$, and let $\overline{W} = \prod [\alpha_i, \beta_i]$. The open cells $W_j = \prod (\alpha_i - \tfrac{1}{j}, \beta_i + \tfrac{1}{j})$ decrease, $m(W_1) < \infty$, and $\bigcap_j W_j = \overline{W}$. By continuity from above and Step 2, $$m(\overline{W}) = \lim_j \operatorname{vol}(W_j) = \operatorname{vol}(W).$$ If some $\alpha_i = \beta_i$, then $0 \le m(W) \le m(\overline{W}) = 0 = \operatorname{vol}(W)$. Otherwise the open cell $\prod (\alpha_i, \beta_i)$ lies inside $W$, which lies inside $\overline{W}$, and both ends have measure $\operatorname{vol}(W)$, so $m(W) = \operatorname{vol}(W)$. This proves (a).
Step 4: the sandwich. $\mathbb{R}^n = \bigcup_j [-j, j]^n$ is σ-compact, so Riesz's theorem for σ-compact spaces says $m$ is regular and gives, for each $E \in \mathcal{M}$, an $F_\sigma$ set $A$ and a $G_\delta$ set $B$ with $A \subseteq E \subseteq B$ and $m(B \setminus A) = 0$. Conversely, suppose such $A$ and $B$ exist. They are Borel, so they're in $\mathcal{M}$. Then $E \setminus A \subseteq B \setminus A$, a set of measure $0$, so $E \setminus A \in \mathcal{M}$ because $m$ is complete, and $E = A \cup (E \setminus A) \in \mathcal{M}$. This proves (b). $\blacksquare$
The heart of it is Step 2: a continuous bump squeezed between a closed cell and an open one. The Riemann integral pins $\Lambda f_j$ between two volumes, and dominated convergence pins $\int f_j\,dm$ to $m(W)$. Here it is in one dimension.
Notice what the proof didn't do: it never assumed $m(E_n) = \operatorname{vol}(E_n)$. That's the conclusion, so it can't also be a step. The only volumes it computed were Riemann integrals. Now the names.
The sets in the σ-algebra $\mathcal{M}$ of the theorem are called Lebesgue measurable.
And the measure gets the name of the man who built the integral on top of it.
The measure $m$ of the theorem is the Lebesgue measure on $\mathbb{R}^n$, written $m_n$ when the dimension matters.
Two quick consequences, both straight from the construction. First, for a continuous $f$ with compact support, $\int f\,dm$ is the Riemann integral, since that's how $\Lambda$ was defined. So the old integral is a special case of the new one. Second, we write $L^1(\mathbb{R}^n)$ for $L^1(m_n)$.
Ends never matter. $[a, b]$, $(a, b)$, $[a, b)$, and $(a, b]$ all have measure $b - a$, because they differ by at most two points and every point is a degenerate cell $[x, x]$ of volume $0$. In $\mathbb{R}^2$, a rectangle with or without its edges has the same area for the same reason: the edges are cells with a flat side.
With finite additivity, that's enough to measure anything built from finitely many cells. Split into disjoint pieces, measure each, add.
Find $m(E)$ for $E = [0, 3) \cup (2, 5] \cup [7, 8]$.
The first two overlap on $(2, 3)$, so together they make $[0, 5]$, of measure $5$. (Adding $3 + 3$ would count the overlap twice.) The piece $[7, 8]$ is disjoint from it and adds $1$. So $m(E) = 5 + 1 = 6$. The gap $(5, 7)$ isn't in $E$, so it doesn't count.
Your turn. Watch for overlaps and gaps, and remember that ends are free.
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Points have measure $0$, and countable additivity takes that much further.
Show that every countable set $E \subseteq \mathbb{R}^n$ has measure $0$. Then find $m(\mathbb{Q} \cap [0, 1])$ and $m([0, 1] \setminus \mathbb{Q})$.
Write $E = \{x_1, x_2, \ldots\}$. Each $\{x_j\}$ is a cell with every side a single point, so $m(\{x_j\}) = 0$, and by countable additivity $m(E) = \sum_j m(\{x_j\}) = 0$.
$\mathbb{Q} \cap [0, 1]$ is countable, so its measure is $0$. It has finite measure, so we may subtract: $m([0, 1] \setminus \mathbb{Q}) = m([0, 1]) - m(\mathbb{Q} \cap [0, 1]) = 1 - 0 = 1$.
So the rationals, dense as they are, take up no room at all, and the irrationals take up all of it. The same goes in the plane for lines: the $x$-axis is the countable union of the flat cells $[n, n + 1] \times [0, 0]$, each of volume $0$. Drop the null pieces, then measure what's left.
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Countable sets are null. Is that the whole story, so that null means countable? The Cantor set says no, and it's the most useful counterexample in the subject.
Let $C_0 = [0, 1]$, and for $n \ge 0$ let $C_{n+1}$ be what's left after removing the open middle third of every interval of $C_n$. The Cantor set is $C = \bigcap_{n=0}^{\infty} C_n$.
So $C_1 = [0, \tfrac{1}{3}] \cup [\tfrac{2}{3}, 1]$, and $C_2$ is four intervals of length $\tfrac{1}{9}$. In general $C_n$ is $2^n$ closed intervals of length $3^{-n}$, so $m(C_n) = (\tfrac{2}{3})^n$. Watch it carve.
The Cantor set: I threw away all of $[0, 1]$'s length and still kept uncountably many points. My fridge works the opposite way.
The removed lengths add to $\tfrac{1}{3} + \tfrac{2}{9} + \tfrac{4}{27} + \cdots = 1$, so all of the length is gone. But the endpoints $0, \tfrac{1}{3}, \tfrac{2}{3}, \tfrac{1}{9}, \ldots$ are never removed, and those are only the beginning. Prove both halves.
Show that $m(C) = 0$, even though $C$ and $\mathbb{R}$ have the same cardinality.
Measure. The sets $C_n$ decrease, $m(C_0) = 1$ is finite, and $m(C_n) = (\tfrac{2}{3})^n$. By continuity from above, $m(C) = \lim_n (\tfrac{2}{3})^n = 0$.
Size. By induction on $n$, the intervals of $C_n$ are exactly the intervals $$\Big[\textstyle\sum_{j=1}^{n} a_j 3^{-j}, \ \sum_{j=1}^{n} a_j 3^{-j} + 3^{-n}\Big] \quad \text{with each } a_j \in \{0, 2\}:$$ removing the middle third of such an interval leaves its left third (next digit $a_{n+1} = 0$) and its right third ($a_{n+1} = 2$). For a sequence $a = (a_j)$ with every $a_j \in \{0, 2\}$, let $\varphi(a) = \sum_{j=1}^{\infty} a_j 3^{-j}$. For each $n$, $\varphi(a)$ lies in the interval of $C_n$ labeled by $a_1, \ldots, a_n$ (the tail adds at most $\sum_{j > n} 2 \cdot 3^{-j} = 3^{-n}$), so $\varphi(a) \in C$.
If $a \neq b$ first differ at the $n$-th digit, then $\varphi(a)$ and $\varphi(b)$ lie in different intervals of $C_n$, which are disjoint, so $\varphi(a) \neq \varphi(b)$. So $\varphi$ is one-to-one, and $C$ is at least as large as the set of $0$–$2$ sequences, which has the cardinality of $\mathcal{P}(\mathbb{N})$, that is, of $\mathbb{R}$. And $C \subseteq \mathbb{R}$, so $C$ and $\mathbb{R}$ have the same cardinality. $\blacksquare$
So "null" and "small" are different ideas: the Cantor set has as many points as $\mathbb{R}$ and no length. Now flip the construction around. Remove less each time, little enough that the removed lengths add up to less than $1$, and the set left over keeps positive measure. It still contains no interval, because its pieces still shrink to nothing.
A constant removal fraction always kills the measure, since $(1 - r)^n \to 0$. A shrinking one doesn't, since the removed lengths form a convergent series. That's the trick behind the next exercise, which sounds impossible: an open set that's everywhere in $[0, 1]$ and yet almost takes up no room.
Let $0 < \varepsilon < 1$. Construct an open set $E \subseteq [0, 1]$ that is dense in $[0, 1]$ (its closure contains $[0, 1]$) with $m(E) = \varepsilon$.
Start with $K_0 = [0, 1]$. At stage $n \ge 1$, $K_{n-1}$ is $2^{n-1}$ closed intervals; remove from the middle of each an open interval of length $\varepsilon \, 2^{1 - 2n}$, and call what's left $K_n$. Stage $n$ removes $2^{n-1} \cdot \varepsilon \, 2^{1-2n} = \varepsilon \, 2^{-n}$ in all.
It fits. After $n - 1$ stages the removed length is $\varepsilon(1 - 2^{1-n})$, so each of the $2^{n-1}$ equal intervals of $K_{n-1}$ has length $L_{n-1} = \big(1 - \varepsilon + \varepsilon \, 2^{1-n}\big) / 2^{n-1}$. The piece we remove is shorter: $\varepsilon \, 2^{1-2n} < L_{n-1}$ means $\varepsilon \, 2^{-n} < 1 - \varepsilon + \varepsilon \, 2^{1-n}$, which holds since $\varepsilon \, 2^{-n} < \varepsilon \, 2^{1-n}$ and $1 - \varepsilon > 0$.
The set. Let $K = \bigcap_n K_n$ (a "fat Cantor set") and $E = [0, 1] \setminus K$, the union of all the removed open intervals. $E$ is a union of open intervals, so it's open, and $E \subseteq [0, 1]$. By continuity from above, $m(K) = \lim_n \big(1 - \varepsilon(1 - 2^{-n})\big) = 1 - \varepsilon$, so $m(E) = 1 - m(K) = \varepsilon$.
Dense. The intervals of $K_n$ have length $L_n \le 2^{-n}$. If an open interval $I \subseteq [0, 1]$ missed $E$, then $I \subseteq K \subseteq K_n$ for every $n$, so $I$ would sit inside one interval of $K_n$ and have length at most $2^{-n}$, for every $n$. That's impossible, so every open interval in $[0, 1]$ meets $E$, and the closure of $E$ contains $[0, 1]$. $\blacksquare$
So "dense" and "big" are also different ideas. Here's a sorting round on what we know so far: which of these sets take up no room?
The Cantor set also settles a question the theorem raised: is $\mathcal{M}$ really bigger than the Borel sets?
Every subset of $C$ is Lebesgue measurable: $m(C) = 0$ and $m$ is complete. $C$ has as many points as $\mathbb{R}$, so there are as many Lebesgue measurable sets as there are subsets of $\mathbb{R}$. But there are only as many Borel sets as real numbers (Rudin proves this with a transfinite count; we take it on faith here). So most Lebesgue measurable sets are not Borel. Don't assume a Lebesgue measurable set is Borel.
Now carve your own. The construction changes, but the bookkeeping is always the same: how many pieces, and how much each one loses.
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Back to the wish list. Sliding a cell gives a cell with the same sides, so $m(W + x) = m(W)$ for every cell. The lemma about dyadic boxes carries that to every open set, and regularity carries it the rest of the way.
For every $E \in \mathcal{M}$ and $x \in \mathbb{R}^n$, the translate $E + x$ is Lebesgue measurable, and $$m(E + x) = m(E).$$
Proof. Fix $x$. The map $y \mapsto y + x$ is a homeomorphism of $\mathbb{R}^n$, so it sends open sets to open sets, and so Borel sets to Borel sets (the sets it sends to Borel sets form a σ-algebra containing the open sets). For Borel $E$ let $\lambda(E) = m(E + x)$. Translates of disjoint sets are disjoint, so $\lambda$ is a positive Borel measure.
For an $n$-cell $W$, $W + x$ is a cell with the same side lengths, so $\lambda(W) = m(W)$. An open set $V$ is a countable disjoint union of dyadic boxes $Q_j$ (the lemma), and then $V + x$ is the disjoint union of the $Q_j + x$, so $$\lambda(V) = \sum_j m(Q_j + x) = \sum_j m(Q_j) = m(V).$$ Also $\lambda(K) = m(K + x) < \infty$ for compact $K$, since $K + x$ is compact. Every open set in $\mathbb{R}^n$ is σ-compact, so $\lambda$ is regular (by the regularity theorem on the Riesz page), and so is $m$. By outer regularity, for every Borel $E$, $$\lambda(E) = \inf\{\lambda(V) : V \supseteq E \text{ open}\} = \inf\{m(V) : V \supseteq E \text{ open}\} = m(E).$$ Now let $E \in \mathcal{M}$, with $A \subseteq E \subseteq B$ as in part (b) of the theorem. Translating keeps $F_\sigma$ and $G_\delta$ sets of the same kind, $A + x \subseteq E + x \subseteq B + x$, and $m((B + x) \setminus (A + x)) = m((B \setminus A) + x) = m(B \setminus A) = 0$. So $E + x \in \mathcal{M}$ by (b), and $m(E + x) = m(A + x) = m(A) = m(E)$. $\blacksquare$
A first use: null sets stay null when you move them, even countably many times.
Let $C$ be the Cantor set. Show that $C + \mathbb{Q} = \{c + q : c \in C, \ q \in \mathbb{Q}\}$ has Lebesgue measure $0$.
$C + \mathbb{Q} = \bigcup_{q \in \mathbb{Q}} (C + q)$, a countable union. Each $C + q$ is measurable with $m(C + q) = m(C) = 0$ by translation invariance. By countable subadditivity, $m(C + \mathbb{Q}) \le \sum_{q} 0 = 0$.
So a set can be dense in $\mathbb{R}$ (every interval contains a rational translate of $0 \in C$) and uncountable, and still be null.
That's the pattern for proving two measures equal: agree on boxes, then on open sets (disjoint boxes), then on Borel sets (regularity), then on everything (the sandwich). The same pattern proves something stronger. Lebesgue measure isn't just a translation-invariant measure, it's essentially the only one.
Let $\mu$ be a positive measure on the Borel sets of $\mathbb{R}^n$ that is translation invariant ($\mu(E + x) = \mu(E)$ for all Borel $E$ and all $x$) and finite on compact sets. Then there is a constant $c$ with $0 \le c < \infty$ such that $\mu(E) = c \, m(E)$ for every Borel set $E$.
Proof. Let $Q_0 = [0, 1)^n$ and $c = \mu(Q_0)$. Then $c \le \mu([0, 1]^n) < \infty$, since $[0, 1]^n$ is compact.
Every level-$j$ dyadic box $Q$ is a translate of $[0, 2^{-j})^n$, and $Q_0$ is the disjoint union of $2^{jn}$ level-$j$ boxes. By translation invariance all of them have the same $\mu$-measure, so $2^{jn} \mu(Q) = c$, that is, $\mu(Q) = 2^{-jn} c = c \, m(Q)$.
Every open set $V$ is a countable disjoint union of dyadic boxes, so $\mu(V) = c \, m(V)$. Every open set in $\mathbb{R}^n$ is σ-compact and $\mu$ is finite on compact sets, so $\mu$ is regular, and so is $m$. For a Borel set $E$, outer regularity gives $$\mu(E) = \inf_{V \supseteq E \text{ open}} \mu(V) = \inf_{V \supseteq E \text{ open}} c \, m(V) = c \, m(E).$$ (If $c = 0$, both sides are $0$, with $0 \cdot \infty = 0$.) $\blacksquare$
In practice the theorem turns "find the measure" into "find one number".
Let $\mu(E) = m(E + 7) + 2\,m(E)$ for Borel $E \subseteq \mathbb{R}$. Show that $\mu = c \cdot m$ and find $c$.
$\mu$ is a positive Borel measure (a sum of measures). It is translation invariant: $\mu(E + x) = m(E + x + 7) + 2\,m(E + x) = m(E + 7) + 2\,m(E) = \mu(E)$. And it is finite on compact sets, since $\mu(K) = 3\,m(K) < \infty$. By the theorem, $\mu = c \cdot m$ with $c = \mu([0, 1)) = 1 + 2 = 3$.
(Here you could see $\mu = 3m$ directly. The theorem earns its keep when $\mu$ comes from somewhere less transparent, like $E \mapsto m(T(E))$ below.)
Both hypotheses are doing work, and the usual suspects each fail one of them.
Check both hypotheses before you conclude $\mu = c \cdot m$. Counting measure on $\mathbb{R}$ is translation invariant (moving a set doesn't change how many points it has), but it gives $[0, 1]$ measure $\infty$, so it isn't $c \cdot m$ for any $c$. The Dirac measure $\delta_0$ is finite on compact sets but not translation invariant: $\delta_0([0, 1]) = 1$ while $\delta_0([0, 1] + 2) = 0$.
Pick a measure and slide a set around. Some measures don't notice, some do, and one doesn't notice for the wrong reason.
When both checks pass, the constant is $c = \mu([0, 1))$, the measure of one unit box. Pick out the multiples of $m$.
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Uniqueness pays off right away. Stretch $\mathbb{R}^n$ by a linear map $T$ and measure the images: $E \mapsto m(T(E))$ is translation invariant (because $T$ is linear), so it must be a constant times $m$. Before the theorem, one fact it needs for the maps that squash space flat.
Show that every proper subspace $H$ of $\mathbb{R}^n$ (a subspace other than $\mathbb{R}^n$ itself) has $m(H) = 0$.
Proof. $H$ lies in a hyperplane $\{y : u \cdot y = 0\}$ for some $u \neq 0$, so it is enough to treat that hyperplane, which we still call $H$. $H$ is closed, so it's Borel. Fix $N$ and let $H_N = H \cap [-N, N]^n$, a compact set. For $t \in \mathbb{R}$, the translate $H_N + t u$ lies in $\{y : u \cdot y = t|u|^2\}$, so translates by different $t$ are disjoint. For $t = 1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots$ they all lie in the bounded cell $[-N - |u|, N + |u|]^n$, of finite measure $M$. By translation invariance and countable additivity, $$\sum_{j=1}^{\infty} m(H_N) = \sum_{j=1}^{\infty} m\big(H_N + \tfrac{1}{j} u\big) \le M < \infty,$$ which forces $m(H_N) = 0$. Then $H = \bigcup_N H_N$ has measure $0$. $\blacksquare$
In particular, every line in $\mathbb{R}^2$ and every plane in $\mathbb{R}^3$ is null (translates of null sets are null). Now the theorem.
Let $T\colon \mathbb{R}^n \to \mathbb{R}^n$ be linear. There is a number $\Delta(T) \in [0, \infty)$ such that for every Borel set $E$, $T(E)$ is Lebesgue measurable and $$m(T(E)) = \Delta(T) \, m(E).$$ In particular, $m(T(E)) = m(E)$ when $T$ is a rotation.
Proof. Suppose first that $T$ has rank less than $n$. Then $T(\mathbb{R}^n)$ is a proper subspace, which is null by the exercise, and $T(E)$ lies inside it. $T(E)$ need not be a Borel set, but $m$ is complete, so $T(E)$ is measurable with $m(T(E)) = 0$, and $\Delta(T) = 0$ works.
Now suppose $T$ has rank $n$. Then $T$ is a linear bijection, hence a homeomorphism of $\mathbb{R}^n$, so it maps Borel sets to Borel sets. Let $\mu(E) = m(T(E))$ for Borel $E$. Since $T$ is one-to-one, it keeps disjoint sets disjoint, so $\mu$ is a positive Borel measure. It is finite on compact sets, since $T(K)$ is compact when $K$ is. And it is translation invariant: since $T$ is linear, $T(E + x) = T(E) + T(x)$, so $$\mu(E + x) = m(T(E) + T(x)) = m(T(E)) = \mu(E).$$ By the uniqueness theorem, $\mu = c \, m$ for a constant $0 \le c < \infty$; let $\Delta(T) = c$.
If $T$ is a rotation, it maps the open unit ball $B$ onto itself, and $0 < m(B) < \infty$. So $m(B) = \Delta(T) \, m(B)$ forces $\Delta(T) = 1$. $\blacksquare$
So every linear map has an exchange rate for volume, and you can read it off any one set: $\Delta(T) = m(T(E)) / m(E)$ for any $E$ with $0 < m(E) < \infty$. The unit box is the natural choice. A shear tilts it into a parallelogram, and cutting and pasting shows the area didn't change.
A shear is just a square leaning on the wall. Same area, more attitude.
That cut-and-paste is the only nontrivial step in computing $\Delta(T)$ for every $T$. The other building blocks are a swap of two coordinates, which doesn't change the box at all, and a stretch of one coordinate, which multiplies its volume by the stretch factor.
For every linear $T\colon \mathbb{R}^n \to \mathbb{R}^n$, $$\Delta(T) = |\det T|.$$
Proof. If $T$ is not invertible, its rank is less than $n$, so $\Delta(T) = 0 = |\det T|$ by the proof of the theorem. So let $T$ be invertible.
Both sides multiply. If $T = T_1 T_2$, then for every Borel $E$, $$m(T(E)) = m(T_1(T_2(E))) = \Delta(T_1) \, m(T_2(E)) = \Delta(T_1) \Delta(T_2) \, m(E),$$ so $\Delta(T_1 T_2) = \Delta(T_1) \Delta(T_2)$, and $|\det(T_1 T_2)| = |\det T_1| \, |\det T_2|$. By Gaussian elimination, every invertible matrix is a product of elementary matrices, one for each elementary row operation: (I) swap coordinates $i$ and $j$; (II) multiply coordinate $i$ by some $\alpha \neq 0$; (III) add coordinate $j$ to coordinate $i$, that is, $x \mapsto x + x_j e_i$ with $i \neq j$. So it's enough to check $\Delta = |\det|$ for these. Take $E = [0, 1)^n$, with $m(E) = 1$.
(I) $T(E) = E$, so $\Delta(T) = 1 = |{-1}| = |\det T|$.
(II) $T(E)$ is a cell with side $|\alpha|$ in coordinate $i$ (the interval $[0, \alpha)$ or $(\alpha, 0]$) and side $1$ in the others, so $\Delta(T) = |\alpha| = |\det T|$.
(III) Here $\det T = 1$. Rename the coordinates so that $i = 1$ and $j = 2$; then $T(x_1, x_2, \ldots, x_n) = (x_1 + x_2, x_2, \ldots, x_n)$ and $$T(E) = \{y : y_2 \le y_1 < y_2 + 1, \ 0 \le y_l < 1 \text{ for } l \ge 2\}.$$ Let $S_1 = T(E) \cap \{y_1 < 1\}$ and $S_2 = T(E) \cap \{y_1 \ge 1\}$. In the $(y_1, y_2)$ coordinates, $S_1$ is $\{y_2 \le y_1 < 1\}$ and $S_2 - e_1$ is $\{0 \le y_1 < y_2\}$ (the other coordinates stay in $[0, 1)$). These are disjoint, and their union is $E$. So $$m(T(E)) = m(S_1) + m(S_2) = m(S_1) + m(S_2 - e_1) = m(E) = 1 = \det T. \qquad \blacksquare$$
On Determinants, $\det T$ was the signed area of the parallelogram spanned by the columns. Measure theory keeps the size and throws away the sign: a flip doesn't make a set's area negative.
Find the area of the parallelogram $P$ with corners $(0, 0)$, $(3, 1)$, $(1, 2)$, and $(4, 3)$.
$P$ is the image of the unit square $[0, 1]^2$ under the linear map $T$ with columns $(3, 1)$ and $(1, 2)$: $T(s, t) = s(3, 1) + t(1, 2)$. Then $\det T = 3 \cdot 2 - 1 \cdot 1 = 5$, so $m_2(P) = |\det T| \, m_2([0, 1]^2) = 5$.
So $m(T(E)) = |\det T| \, m(E)$ for every Borel set, not just for parallelograms. Drag the columns of a matrix and watch a blob and its image.
The ratio never depends on the blob, only on the matrix. Dilations are the most common case in practice.
Let $E \subseteq \mathbb{R}^n$ be a Borel set, $r > 0$, and $x \in \mathbb{R}^n$. Find $m(rE + x)$, where $rE = \{ry : y \in E\}$.
$y \mapsto ry$ is the linear map $rI$, with $\det(rI) = r^n$. So $m(rE) = r^n m(E)$, and translating by $x$ changes nothing: $m(rE + x) = r^n \, m(E)$. Doubling a set in $\mathbb{R}^3$ multiplies its volume by $8$, not $2$.
Now compute some exchange rates. The sign of the determinant is where most of the trouble lives.
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We built $m$ on a σ-algebra $\mathcal{M}$, not on every subset of $\mathbb{R}^n$. Was that just caution? No. Translation invariance is so strong that no translation-invariant measure with $0 < \mu([0, 1)) < \infty$ can measure every set, and the counterexample fits in a paragraph, once you allow yourself the Axiom of Choice.
The Vitali set: brought to you by the Axiom of Choice, which never has to show its work.
Let $\mathcal{M}$ be a σ-algebra on $\mathbb{R}$ that contains $[0, 1)$ and is closed under translation ($E + x \in \mathcal{M}$ whenever $E \in \mathcal{M}$), and let $\lambda$ be a translation-invariant positive measure on $\mathcal{M}$ with $0 < \lambda([0, 1)) < \infty$. Then some $E \subseteq [0, 1)$ is not in $\mathcal{M}$. In particular, not every subset of $\mathbb{R}$ is Lebesgue measurable.
Proof. On $[0, 1)$, let $x \sim y$ when $x - y \in \mathbb{Q}$. This is an equivalence relation. By the Axiom of Choice, there is a set $E \subseteq [0, 1)$ containing exactly one point from each equivalence class. Let $R = \mathbb{Q} \cap (-1, 1)$, a countably infinite set.
The translates $E + r$, $r \in R$, are pairwise disjoint. Suppose $y + r = z + s$ with $y, z \in E$ and $r, s \in R$. Then $y - z = s - r \in \mathbb{Q}$, so $y \sim z$. Since $E$ has only one point from each class, $y = z$, and then $r = s$.
They cover $[0, 1)$. Let $x \in [0, 1)$. Its class has a representative $y \in E$, and $r = x - y$ is rational with $|r| < 1$, since $x, y \in [0, 1)$. So $x = y + r \in E + r$ with $r \in R$.
They stay inside $[-1, 2)$, since $E \subseteq [0, 1)$ and $|r| < 1$.
Now suppose $E \in \mathcal{M}$. Then every $E + r \in \mathcal{M}$ with $\lambda(E + r) = \lambda(E)$. Also $[-1, 2) = ([0, 1) - 1) \cup [0, 1) \cup ([0, 1) + 1)$ is in $\mathcal{M}$, with $\lambda([-1, 2)) = 3 \lambda([0, 1))$. By the three facts, $$\lambda([0, 1)) \le \sum_{r \in R} \lambda(E + r) = \sum_{r \in R} \lambda(E) \le \lambda([-1, 2)) = 3 \lambda([0, 1)) < \infty.$$ A sum of infinitely many copies of $\lambda(E)$ is finite only if $\lambda(E) = 0$. But then the left inequality says $\lambda([0, 1)) \le 0$, contradicting $\lambda([0, 1)) > 0$. So $E \notin \mathcal{M}$.
Lebesgue measure satisfies the hypotheses: $\mathcal{M}$ is closed under translation and $m([0, 1)) = 1$. $\blacksquare$
The proof is a squeeze with nowhere to go. Infinitely many disjoint copies of $E$ fit inside a set of measure $3$, so $E$ can't have positive measure. But they also cover a set of measure $1$, so $E$ can't have measure $0$. Here are both readouts at once.
Pick your own value for $m(E)$ and see which wall you hit.
Every hypothesis of the proposition was used. Drop one and the conclusion can fail.
Counting measure is translation invariant and is defined on every subset of $\mathbb{R}$. Why doesn't that contradict the proposition?
The proposition needs $0 < \lambda([0, 1)) < \infty$, and counting measure gives $[0, 1)$ measure $\infty$. In the proof, the bound $\sum_{r} \lambda(E) \le 3\lambda([0, 1))$ becomes $\le \infty$, which forces nothing: the Vitali set just has counting measure $\infty$, and nothing breaks.
There's a second road to Lebesgue measure that doesn't go through Riesz: cover a set by intervals in every possible way and take the smallest total length. That "outer measure" is defined for every set, and the Vitali set shows why it can't be countably additive on all of them. Carathéodory's test picks out the sets where it is. That's the next road on Outer Measure & Carathéodory.
So $\mathcal{M}$ is a genuine restriction, and it has to be. The proofs on this page all come down to a handful of moves: disjoint boxes, regularity, the sandwich, and counting translates. Fill in the missing move.
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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; the types whose choices are measures or proof lines will say they need their choices.
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Last, the Spicy Bonus. Vitali found one bad subset of $[0, 1)$. In fact, bad sets are everywhere: you can't have positive measure without containing one.
Prove that every Lebesgue measurable set $A \subseteq \mathbb{R}$ with $m(A) > 0$ contains a subset that is not Lebesgue measurable.
Proof. Let $E \subseteq [0, 1)$ be the Vitali set from the proposition: one point from each class of the relation $x - y \in \mathbb{Q}$.
Step 1: the translates of $E$ by all rationals partition $\mathbb{R}$. They're disjoint by the same argument as in the proposition (that argument never used $|r| < 1$). For $x \in \mathbb{R}$, the number $x - \lfloor x \rfloor$ lies in $[0, 1)$, so it is equivalent to some $y \in E$, and then $x - y = (x - \lfloor x \rfloor - y) + \lfloor x \rfloor$ is rational. So $x \in E + r$ for some $r \in \mathbb{Q}$.
Step 2: every measurable subset of a translate $E + r$ is null. Let $B \subseteq E + r$ be measurable. Then $B' = B - r \subseteq E$ is measurable, with $m(B') = m(B)$. The sets $B' + q$ for $q \in \mathbb{Q} \cap (-1, 1)$ are disjoint (they sit inside the disjoint sets $E + q$), measurable, all of measure $m(B)$, and all inside $[-1, 2)$. So $$\sum_{q \in \mathbb{Q} \cap (-1, 1)} m(B) \le m([-1, 2)) = 3,$$ and infinitely many equal terms with a finite sum are $0$: $m(B) = 0$.
Step 3. By Step 1, $A = \bigcup_{r \in \mathbb{Q}} \big(A \cap (E + r)\big)$, a countable union. If every $A \cap (E + r)$ were measurable, each would be null by Step 2, and then $m(A) \le \sum_r m(A \cap (E + r)) = 0$, contradicting $m(A) > 0$. So some $A \cap (E + r)$ is a subset of $A$ that is not Lebesgue measurable. $\blacksquare$