On Hahn–Banach, the dual $X^*$ of a normed space was the set of bounded linear functionals on it, with $\|\Lambda\| = \sup\{|\Lambda(x)| : \|x\| \le 1\}$. Hahn–Banach said $X^*$ is big. It didn't say what's in it.
This page answers that question for the two kinds of spaces this course cares about most. On $L^p(\mu)$, there's an obvious way to build a functional: pick a function $g$, multiply, integrate. On $C_0(X)$, there's another: pick a measure $\mu$ and integrate against it. The question both times is whether the obvious functionals are all of them.
Both answers are yes, with the right hypotheses, and both proofs run on the same engine as the last page. Feed a functional the characteristic function of a set and you get a number for every set. That's a measure, and the Radon–Nikodym theorem turns a measure into a function.
Throughout, $\mu$ is a positive measure on $(X, \mathcal{M})$, $1 \le p \le \infty$, and $q$ is the conjugate exponent, $\frac{1}{p} + \frac{1}{q} = 1$, with $1$ and $\infty$ a pair.
Start with the functionals we can write down. Fix $g \in L^q(\mu)$ and set $$\Lambda_g(f) = \int_X fg \, d\mu \qquad (f \in L^p(\mu)).$$ Is that even defined? That's exactly what Hölder's inequality is for.
On The Lᵖ Spaces: if $f \in L^p(\mu)$ and $g \in L^q(\mu)$, then $fg \in L^1(\mu)$ and $\|fg\|_1 \le \|f\|_p \|g\|_q$, for every $1 \le p \le \infty$.
So $|\Lambda_g(f)| \le \|f\|_p \|g\|_q$, and $\Lambda_g$ is a bounded functional with $\|\Lambda_g\| \le \|g\|_q$. The interesting half is the other inequality: is there an $f$ that makes Hölder an equality? Usually, yes.
Let $g \in L^q(\mu)$. Then $\Lambda_g(f) = \int_X fg \, d\mu$ is a bounded linear functional on $L^p(\mu)$ with $\|\Lambda_g\| \le \|g\|_q$. Equality $\|\Lambda_g\| = \|g\|_q$ holds if $1 < p \le \infty$, and also if $p = 1$ and $\mu$ is σ-finite.
Proof. Linearity is linearity of the integral, and Hölder gives $|\Lambda_g(f)| \le \int_X |fg| \, d\mu \le \|f\|_p \|g\|_q$. So $\|\Lambda_g\| \le \|g\|_q$. For the reverse inequality we may assume $\|g\|_q > 0$. Let $\alpha = \overline{g}/|g|$ where $g \ne 0$ and $\alpha = 1$ where $g = 0$. Then $\alpha$ is measurable, $|\alpha| = 1$, and $\alpha g = |g|$.
$1 < p < \infty$. Let $f = \alpha |g|^{q - 1}$. Since $\frac{1}{p} + \frac{1}{q} = 1$ means $p(q - 1) = q$, we get $|f|^p = |g|^{p(q - 1)} = |g|^q$, so $\|f\|_p = \|g\|_q^{q/p}$. And $fg = |g|^q$, so $\Lambda_g(f) = \|g\|_q^q$. Therefore $$\|\Lambda_g\| \ge \frac{|\Lambda_g(f)|}{\|f\|_p} = \frac{\|g\|_q^q}{\|g\|_q^{q/p}} = \|g\|_q^{q - q/p} = \|g\|_q.$$
$p = \infty$, $q = 1$. Let $f = \alpha$. Then $\|f\|_\infty = 1$ and $\Lambda_g(f) = \int_X |g| \, d\mu = \|g\|_1$.
$p = 1$, $q = \infty$, $\mu$ σ-finite. Let $0 < t < \|g\|_\infty$. By the definition of the essential sup, $A = \{|g| > t\}$ has $\mu(A) > 0$. Write $X = \bigcup_n X_n$ with $\mu(X_n) < \infty$. Then $\mu(A \cap X_n) > 0$ for some $n$; call that set $E$, so $0 < \mu(E) < \infty$. Let $f = \alpha \chi_E / \mu(E)$. Then $\|f\|_1 = 1$ and $$\Lambda_g(f) = \frac{1}{\mu(E)} \int_E |g| \, d\mu \ge t.$$ So $\|\Lambda_g\| \ge t$ for every $t < \|g\|_\infty$, and $\|\Lambda_g\| \ge \|g\|_\infty$. $\blacksquare$
The $f$ in that proof is the whole trick: it lines up with $g$ in direction ($\alpha g = |g|$) and spends its size where $|g|$ is big. Normalized, it gets a name.
Let $g \in L^q(\mu)$. A norming function for $\Lambda_g$ on $L^p(\mu)$ is an $f \in L^p(\mu)$ with $\|f\|_p = 1$ and $\Lambda_g(f) = \|\Lambda_g\|$.
For $1 < p < \infty$ the proof hands us one: $f = \overline{g}|g|^{q - 2}/\|g\|_q^{q - 1}$ (and $0$ where $g = 0$). Its $L^p$ norm is $\|g\|_q^{q/p}/\|g\|_q^{q - 1} = 1$, since $\frac{q}{p} = q - 1$. Let's see one in action, at three different values of $p$.
On $[0, 5]$ with Lebesgue measure, let $g = 3$ on $[0, 1)$ and $g = -2$ on $[1, 5]$. Find $\|\Lambda_g\|$ and a norming function on $L^2$, on $L^\infty$, and on $L^1$.
$p = 2$, so $q = 2$. $\|g\|_2^2 = 3^2 \cdot 1 + 2^2 \cdot 4 = 25$, so $\|\Lambda_g\| = 5$. The norming function is $f = g/5$: $\frac{3}{5}$ on $[0, 1)$ and $-\frac{2}{5}$ on $[1, 5]$. Check: $\|f\|_2^2 = \frac{9}{25} + 4 \cdot \frac{4}{25} = 1$, and $\Lambda_g(f) = \frac{9}{5} + 4 \cdot \frac{4}{5} = 5$.
$p = \infty$, so $q = 1$. $\|\Lambda_g\| = \|g\|_1 = 3 + 2 \cdot 4 = 11$. The norming function is the sign of $g$: $f = 1$ on $[0, 1)$, $f = -1$ on $[1, 5]$, and then $fg = |g|$ integrates to $11$.
$p = 1$, so $q = \infty$. $\|\Lambda_g\| = \|g\|_\infty = 3$. Spend all of $f$'s mass where $|g|$ is biggest: $f = \chi_{[0, 1)}$ has $\|f\|_1 = 1$ and $\Lambda_g(f) = 3$.
Three exponents, three answers: $5$, $11$, $3$. As $p$ grows, the best $f$ spreads out from the peak of $|g|$ to all of $g$'s support.
That spreading is worth watching. Below, one step function $g$ stays put while $p$ slides from near $1$ up to $\infty$, and the norming function morphs from a spike on the tallest step to a flat copy of $g$'s signs.
Now pick your own $g$. Drag the steps, slide $p$, and watch the norming function and the readout $\Lambda_g(f) = \|g\|_q$.
The proof found the best $f$ with a formula. Without the formula, it's a game: shape $f$ yourself, keep $\|f\|_p = 1$, and push $\Lambda_g(f)$ as high as it goes.
You can't beat $\|g\|_q$; Hölder says so. You can only tie it. One case in the Proposition needed an extra hypothesis, and it's not decoration.
For $p = 1$, equality needs σ-finiteness. Let $X = \{a, b\}$ with $\mu(\{a\}) = 1$ and $\mu(\{b\}) = \infty$, and let $g = \chi_{\{b\}}$. Then $\|g\|_\infty = 1$, because $\{b\}$ has positive measure. But every $f \in L^1(\mu)$ has $f(b) = 0$ (otherwise $\int |f| \, d\mu = \infty$), so $\Lambda_g(f) = \int f\chi_{\{b\}} \, d\mu = 0$ for every $f$. So $\|\Lambda_g\| = 0 \ne 1 = \|g\|_\infty$.
Your turn. The norm of $\Lambda_g$ is a norm of $g$, but in the other exponent.
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And the norming function itself, at one point. Remember it has to line up with $g$'s sign and have $L^p$ norm exactly $1$.
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So $g \mapsto \Lambda_g$ puts a copy of $L^q$ inside $(L^p)^*$, without stretching anything. The hard direction is the converse: start from an arbitrary $\Lambda$ and find its $g$. The proof is easiest when $\mu(X) < \infty$, because then every $\chi_E$ is in $L^p$. A trick moves any σ-finite measure to a finite one without changing $L^p$ in any way that matters.
On The Riesz Representation Theorem, $\mu$ was σ-finite if $X$ is a countable union of sets of finite measure. Lebesgue measure on $\mathbb{R}^n$ is the standard example.
If $\mu$ is a σ-finite positive measure on $(X, \mathcal{M})$, there is $w \in L^1(\mu)$ with $0 < w(x) < 1$ for every $x \in X$.
Proof. Write $X = \bigcup_n A_n$ with $\mu(A_n) < \infty$, and make the pieces disjoint: $X_1 = A_1$, $X_n = A_n \setminus (A_1 \cup \cdots \cup A_{n-1})$. Each $X_n$ is measurable with $\mu(X_n) < \infty$. Define $$w = \sum_{n=1}^\infty \frac{2^{-n}}{1 + \mu(X_n)} \, \chi_{X_n}.$$ Each $x$ lies in exactly one $X_n$, so $0 < w(x) \le \frac{1}{2} < 1$. By monotone convergence, $$\int_X w \, d\mu = \sum_{n=1}^\infty 2^{-n} \frac{\mu(X_n)}{1 + \mu(X_n)} < \sum_{n=1}^\infty 2^{-n} = 1. \qquad \blacksquare$$
The weight gets small on the pieces far out, fast enough to make the total finite, but it never touches $0$. Here it is for Lebesgue measure on $\mathbb{R}$.
Build the weight $w$ of the lemma for Lebesgue measure $m$ on $\mathbb{R}$, using the pieces $X_n = \{x : n - 1 \le |x| < n\}$, and compute $\int_{\mathbb{R}} w \, dm$.
The pieces are disjoint and cover $\mathbb{R}$: $X_1 = (-1, 1)$, $X_2 = (-2, -1] \cup [1, 2)$, and so on. Each has $m(X_n) = 2$, so $$w = \frac{2^{-n}}{1 + 2} = \frac{2^{-n}}{3} \quad \text{on } X_n.$$ So $w = \frac{1}{6}$ on $(-1, 1)$, $\frac{1}{12}$ on the next two unit intervals, and so on. Then $\int_{\mathbb{R}} w \, dm = \sum_n 2 \cdot \frac{2^{-n}}{3} = \frac{2}{3}$.
Watch the weight go down piece by piece while the running total of its integral stays under $1$.
Now use $w$ as a density. The new measure $d\tilde{\mu} = w \, d\mu$ is finite, and it's interchangeable with $\mu$ as far as $L^p$ is concerned.
Let $\mu$ be σ-finite, let $w$ be as in the last lemma, and let $\tilde{\mu}(E) = \int_E w \, d\mu$. Then:
Proof. (a) $\tilde{\mu}$ is a positive measure by the density theorem on The Lebesgue Integral, which also gives $\int_X \varphi \, d\tilde{\mu} = \int_X \varphi w \, d\mu$ for measurable $\varphi \ge 0$. And $\tilde{\mu}(X) = \int_X w \, d\mu < 1$.
(b) If $\mu(E) = 0$, then $\int_E w \, d\mu = 0$. Conversely, if $\int_E w \, d\mu = 0$, then $w\chi_E = 0$ $\mu$-a.e.; since $w > 0$ everywhere, $\chi_E = 0$ $\mu$-a.e., that is, $\mu(E) = 0$.
(c) For measurable $\tilde{f}$, $$\int_X |w^{1/p}\tilde{f}|^p \, d\mu = \int_X |\tilde{f}|^p \, w \, d\mu = \int_X |\tilde{f}|^p \, d\tilde{\mu},$$ so $\|\iota(\tilde{f})\|_{L^p(\mu)} = \|\tilde{f}\|_{L^p(\tilde{\mu})}$. By (b), functions equal $\tilde{\mu}$-a.e. are equal $\mu$-a.e., so $\iota$ is well defined on classes; it is clearly linear. It is onto: given $f \in L^p(\mu)$, the same computation shows $\tilde{f} = w^{-1/p} f \in L^p(\tilde{\mu})$, and $\iota(\tilde{f}) = f$.
(d) The essential sup of $|f|$ is decided by which sets $\{|f| > t\}$ are null, and by (b) the two measures have the same null sets. $\blacksquare$
So for $p < \infty$ the spaces $L^p(\mu)$ and $L^p(\tilde{\mu})$ are the same Banach space in disguise, and so are their duals. Anything we prove about the dual for finite measures transfers to σ-finite ones.
The weight does more than the proof needs. It actually characterizes σ-finiteness, which makes for a short exercise.
Prove the converse of the first lemma: if there is $w \in L^1(\mu)$ with $w(x) > 0$ for every $x$, then $\mu$ is σ-finite.
Proof. Let $E_n = \{w > \frac{1}{n}\}$, a measurable set. Since $w(x) > 0$ for every $x$, every $x$ has $w(x) > \frac{1}{n}$ for some $n$, so $X = \bigcup_n E_n$. And $$\frac{1}{n} \mu(E_n) \le \int_{E_n} w \, d\mu \le \int_X w \, d\mu < \infty,$$ so $\mu(E_n) \le n \|w\|_1 < \infty$. So $X$ is a countable union of sets of finite measure. $\blacksquare$
Now the main event. Take any bounded $\Lambda$ on $L^p(\mu)$ with $\mu$ finite. Each set $E$ has $\chi_E \in L^p$, so $\lambda(E) = \Lambda(\chi_E)$ is a number. That set function turns out to be a complex measure, and it ignores $\mu$-null sets. Here is the idea as a picture.
The slope of the curve $E \mapsto \lambda(E)$ is the function we're after. Turning "a measure that ignores $\mu$-null sets" into "a density" is precisely the last page's theorem.
On The Radon–Nikodym Theorem: if $\mu$ is σ-finite and the complex measure $\lambda$ has $\lambda \ll \mu$, then $\lambda(E) = \int_E g \, d\mu$ for a unique $g \in L^1(\mu)$.
Let $1 \le p < \infty$, let $\mu$ be a σ-finite positive measure, and let $\Lambda$ be a bounded linear functional on $L^p(\mu)$. Then there is a unique $g \in L^q(\mu)$ such that $$\Lambda(f) = \int_X fg \, d\mu \quad \text{for every } f \in L^p(\mu).$$ Moreover, $\|\Lambda\| = \|g\|_q$. So $g \mapsto \Lambda_g$ is an isometric isomorphism of $L^q(\mu)$ onto $L^p(\mu)^*$: $L^p(\mu)^* \cong L^q(\mu)$ isometrically.
Proof. Uniqueness and the norm. Since $p < \infty$, either $p > 1$ or $p = 1$ with $\mu$ σ-finite, so the Proposition gives $\|\Lambda_h\| = \|h\|_q$ for every $h \in L^q(\mu)$. If $g$ and $g'$ both represent $\Lambda$, then $\Lambda_{g - g'} = 0$, so $\|g - g'\|_q = 0$ and $g = g'$ in $L^q(\mu)$. Once $g$ exists, $\|\Lambda\| = \|\Lambda_g\| = \|g\|_q$. So everything rests on existence.
Step 1: $\Lambda$ makes a measure ($\mu$ finite). Assume first that $\mu(X) < \infty$, and define $\lambda(E) = \Lambda(\chi_E)$ for $E \in \mathcal{M}$. By linearity, $\lambda$ is finitely additive. If $E = \bigcup_n E_n$ is a measurable partition and $A_N = E_1 \cup \cdots \cup E_N$, then $$\|\chi_E - \chi_{A_N}\|_p = \mu(E \setminus A_N)^{1/p} \to 0,$$ by continuity from above ($\mu$ is finite) and because $p < \infty$. Since $\Lambda$ is continuous, $\sum_{n \le N} \lambda(E_n) = \lambda(A_N) \to \lambda(E)$. So $\lambda$ is a complex measure. If $\mu(E) = 0$, then $\chi_E = 0$ in $L^p(\mu)$ and $\lambda(E) = 0$: $\lambda \ll \mu$.
Step 2: the density. By the Radon–Nikodym theorem there is $g \in L^1(\mu)$ with $\Lambda(\chi_E) = \lambda(E) = \int_E g \, d\mu$ for every $E$. By linearity, $\Lambda(s) = \int_X sg \, d\mu$ for every simple measurable $s$. Now let $f \in L^\infty(\mu)$; changing it on a null set, take $f$ bounded. There are simple $s_n$ with $\|s_n - f\|_\infty \to 0$ (the staircases on Upper Limits & Simple Functions, applied to the real and imaginary parts, converge uniformly on a bounded function). Then $\|s_n - f\|_p \le \|s_n - f\|_\infty \, \mu(X)^{1/p} \to 0$, so $\Lambda(s_n) \to \Lambda(f)$; and $$\left|\int_X s_n g \, d\mu - \int_X fg \, d\mu\right| \le \|s_n - f\|_\infty \|g\|_1 \to 0.$$ So $$\Lambda(f) = \int_X fg \, d\mu \quad \text{for every } f \in L^\infty(\mu). \tag{$*$}$$
Step 3: $g \in L^\infty$ when $p = 1$. For $E$ with $\mu(E) > 0$, $$\left|\int_E g \, d\mu\right| = |\Lambda(\chi_E)| \le \|\Lambda\| \, \|\chi_E\|_1 = \|\Lambda\| \, \mu(E).$$ So every average $\frac{1}{\mu(E)}\int_E g \, d\mu$ lies in the closed disk of radius $\|\Lambda\|$, and by the averages theorem, $|g| \le \|\Lambda\|$ $\mu$-a.e. So $g \in L^\infty(\mu)$.
Step 4: $g \in L^q$ when $1 < p < \infty$. Let $E_n = \{|g| \le n\}$, take $\alpha$ with $|\alpha| = 1$ and $\alpha g = |g|$, and set $f_n = \chi_{E_n} \alpha |g|^{q - 1}$. Each $f_n$ is bounded, so $(*)$ applies: $$\int_{E_n} |g|^q \, d\mu = \Lambda(f_n) \le \|\Lambda\| \, \|f_n\|_p = \|\Lambda\| \left(\int_{E_n} |g|^q \, d\mu\right)^{1/p},$$ using $|f_n|^p = \chi_{E_n}|g|^q$. The integral is at most $n^q \mu(X) < \infty$. If it is positive, divide by its $\frac{1}{p}$-th power: since $1 - \frac{1}{p} = \frac{1}{q}$, $$\left(\int_{E_n} |g|^q \, d\mu\right)^{1/q} \le \|\Lambda\|,$$ and that holds trivially when the integral is $0$. The sets $E_n$ increase to $X$, because $g$ is complex-valued, so monotone convergence gives $\|g\|_q \le \|\Lambda\|$.
Step 5: all of $L^p$. Now $g \in L^q(\mu)$, so $\Lambda_g$ is bounded on $L^p(\mu)$, by the Proposition. The two bounded functionals $\Lambda$ and $\Lambda_g$ agree on $L^\infty(\mu)$ by $(*)$, and $L^\infty(\mu)$ contains the simple functions, which are dense in $L^p(\mu)$ for $p < \infty$. Continuous maps that agree on a dense set agree everywhere: $\Lambda = \Lambda_g$. That finishes the finite case.
Step 6: σ-finite $\mu$. Take $w$ and $d\tilde{\mu} = w \, d\mu$ from the lemmas, with the isometry $\iota$ of $L^p(\tilde{\mu})$ onto $L^p(\mu)$. Then $\tilde{\Lambda} = \Lambda \circ \iota$ is a bounded functional on $L^p(\tilde{\mu})$, and $\tilde{\mu}$ is finite, so Steps 1–5 give $\tilde{g} \in L^q(\tilde{\mu})$ with $\tilde{\Lambda}(\tilde{f}) = \int_X \tilde{f}\tilde{g} \, d\tilde{\mu}$. Put $g = w^{1/q}\tilde{g}$ (so $g = \tilde{g}$ when $p = 1$, $q = \infty$). For $f \in L^p(\mu)$, $$\Lambda(f) = \tilde{\Lambda}(w^{-1/p} f) = \int_X w^{-1/p} f \, \tilde{g} \, w \, d\mu = \int_X f \, w^{1 - 1/p} \tilde{g} \, d\mu = \int_X fg \, d\mu,$$ since $1 - \frac{1}{p} = \frac{1}{q}$. (The middle step is $\int \varphi \, d\tilde{\mu} = \int \varphi w \, d\mu$, applied to the real and imaginary parts of the integrable function $w^{-1/p} f \tilde{g}$.) Finally $g \in L^q(\mu)$: if $q < \infty$, $\int_X |g|^q \, d\mu = \int_X |\tilde{g}|^q w \, d\mu = \int_X |\tilde{g}|^q \, d\tilde{\mu} < \infty$; if $q = \infty$, $g = \tilde{g}$ and $L^\infty(\tilde{\mu}) = L^\infty(\mu)$. $\blacksquare$
All that machinery, and every functional on $L^p$ turns out to be "multiply by $g$ and integrate." It's a dot product in a trench coat.
Read the proof's Step 4 again: before it, we only knew $g \in L^1$. The norming functions of the truncations $g\chi_{E_n}$ squeeze out a bound on $\|g\chi_{E_n}\|_q$ that doesn't depend on $n$, and monotone convergence does the rest. Here it is for $g(x) = x^{-1/4}$ on $(0, 1]$ and $p = q = 2$, where $\|g\|_2 = \sqrt{2}$.
A side remark: for $1 < p < \infty$ the σ-finiteness hypothesis can be dropped entirely (the proof uses a different idea, the uniform convexity of $L^p$, and isn't on this page). For $p = 1$ it can't, as the next section shows. First, a computation, where the proof's recipe does all the work.
Let $X = \{1, 2, 3\}$ with $\mu(\{1\}) = 2$, $\mu(\{2\}) = 1$, $\mu(\{3\}) = 4$, and let $\Lambda(f) = 6f(1) - f(2) + 2f(3)$ on $L^2(\mu)$. Find the $g$ that represents $\Lambda$, and $\|\Lambda\|$.
Feed $\Lambda$ the characteristic function of each point. $\Lambda(\chi_{\{1\}}) = 6$, and $\int \chi_{\{1\}} g \, d\mu = g(1)\mu(\{1\}) = 2g(1)$, so $g(1) = 3$. Likewise $g(2) = \frac{-1}{1} = -1$ and $g(3) = \frac{2}{4} = \frac{1}{2}$. Check: $\int fg \, d\mu = 3 \cdot f(1) \cdot 2 - f(2) \cdot 1 + \frac{1}{2} f(3) \cdot 4 = \Lambda(f)$.
Since $p = 2$, $q = 2$ and $$\|\Lambda\| = \|g\|_2 = \left(2 \cdot 3^2 + 1 \cdot 1^2 + 4 \cdot \left(\tfrac{1}{2}\right)^2\right)^{1/2} = \sqrt{20} = 2\sqrt{5}.$$ The tempting wrong answer is $\sqrt{6^2 + 1^2 + 2^2} = \sqrt{41}$, which forgets that the coefficients are $g(k)\mu(\{k\})$, not $g(k)$.
Now find a representative yourself. Divide by what $\mu$ puts at the point, not by anything else.
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In that question the formula for $\Lambda$ gave $g$ away. Step 1 of the proof says you don't need the formula: probing with characteristic functions is enough. Below, $\Lambda$ is a black box on $L^2[0, 1]$. Drag a window $E$ and read $\Lambda(\chi_E)$.
The theorem has two hypotheses, $p < \infty$ and σ-finiteness, and each one is load-bearing. Start with $p = \infty$. Every $g \in L^1$ gives a functional $\Lambda_g$ on $L^\infty$ with $\|\Lambda_g\| = \|g\|_1$, by the Proposition. So $L^1$ sits inside $(L^\infty)^*$ isometrically. It just isn't all of it.
$L^\infty$ is the dual of $L^1$ (σ-finite, of course), but $L^1$ is not the dual of $L^\infty$. Duality is not a two-way street.
The proof breaks at Step 1: with $p = \infty$, $\|\chi_E - \chi_{A_N}\|_\infty$ is $1$ until the leftover set is null, so $\lambda$ needn't be countably additive. The functional that exploits this comes from Hahn–Banach.
With counting measure on $\mathbb{N}$, show that there is a bounded linear functional $\Lambda$ on $\ell^\infty$ that is not of the form $\Lambda(x) = \sum_n x_n y_n$ for any $y \in \ell^1$.
Proof. Let $c \subseteq \ell^\infty$ be the subspace of convergent sequences, and let $\lambda(x) = \lim_n x_n$ on $c$. It is linear, and $|\lambda(x)| \le \sup_n |x_n| = \|x\|_\infty$, so $\|\lambda\| \le 1$. By Hahn–Banach, $\lambda$ extends to a bounded linear functional $\Lambda$ on $\ell^\infty$.
Suppose $\Lambda(x) = \sum_n x_n y_n$ for some $y \in \ell^1$. The unit vector $e_k$ converges to $0$, so $y_k = \Lambda(e_k) = \lambda(e_k) = 0$ for every $k$. Then $y = 0$ and $\Lambda = 0$. But $(1, 1, 1, \ldots) \in c$ has $\Lambda(1, 1, 1, \ldots) = \lim 1 = 1$. Contradiction. $\blacksquare$
Now drop σ-finiteness and keep $p = 1$. The smallest counterexample has two points, one of them infinitely heavy. Watch what $\Lambda_g$ sees and what it doesn't.
A point of infinite mass is a black hole for $L^1$: anything integrable has to be zero there.
That picture is a full proof waiting to be written down.
Let $X = \{a, b\}$, with every subset measurable, $\mu(\{a\}) = 1$, and $\mu(\{b\}) = \mu(X) = \infty$. Is $L^\infty(\mu)$ the dual of $L^1(\mu)$?
No, not even as vector spaces. If $f \in L^1(\mu)$, then $|f(a)| \cdot 1 + |f(b)| \cdot \infty < \infty$, which forces $f(b) = 0$. So $L^1(\mu) = \{f : f(b) = 0\}$ is one-dimensional, with $\|f\|_1 = |f(a)|$, and its dual is one-dimensional too: every functional is $f \mapsto c \, f(a)$.
On the other hand, the only $\mu$-null set is $\emptyset$, so $L^\infty(\mu)$ is every function on $X$, with $\|g\|_\infty = \max\{|g(a)|, |g(b)|\}$. That space is two-dimensional, and a two-dimensional space is not isomorphic to a one-dimensional one, isometrically or otherwise. Concretely, the map $g \mapsto \Lambda_g$ is onto $L^1(\mu)^*$ but not one-to-one: $g = \chi_{\{b\}}$ has $\Lambda_g = 0$. $\blacksquare$
There, $L^\infty$ was too big. In the next example it's too small: a perfectly good functional on $L^1$ wants a $g$ that isn't measurable.
Let $\mathcal{M}$ be the σ-algebra of subsets $E \subseteq [0, 1]$ such that $E$ or $[0, 1] \setminus E$ is countable, let $\mu$ be counting measure on $\mathcal{M}$, and let $g(x) = x$. Show that $g$ is not $\mathcal{M}$-measurable, but $\Lambda(f) = \int fg \, d\mu$ makes sense for every $f \in L^1(\mu)$ and is a bounded linear functional that is not $\Lambda_h$ for any $h \in L^\infty(\mu)$.
Proof. $g$ is not measurable. $g^{-1}((0, \frac{1}{2})) = (0, \frac{1}{2})$ is uncountable, and so is its complement $\{0\} \cup [\frac{1}{2}, 1]$. So it is not in $\mathcal{M}$.
$fg$ is integrable. Let $f \in L^1(\mu)$ and $F_n = \{|f| \ge \frac{1}{n}\} \in \mathcal{M}$. By Markov's inequality, $\mu(F_n) \le n\|f\|_1 < \infty$, so each $F_n$ is finite, and $F = \{f \ne 0\} = \bigcup_n F_n$ is countable. The function $fg$ vanishes off $F$. For open $V \subseteq \mathbb{C}$, the set $(fg)^{-1}(V)$ is a countable subset of $F$ together with either nothing (if $0 \notin V$) or all of $[0, 1] \setminus F$ (if $0 \in V$). Either way it is countable or co-countable, so $fg$ is measurable. And $\int |fg| \, d\mu = \sum_{x \in F} x|f(x)| \le \sum_{x \in F} |f(x)| = \|f\|_1$.
$\Lambda$ is bounded. It is linear, and $|\Lambda(f)| \le \|f\|_1$, so $\|\Lambda\| \le 1$.
$\Lambda$ is no $\Lambda_h$. Suppose $h$ is $\mathcal{M}$-measurable and bounded with $\Lambda(f) = \int fh \, d\mu$ for every $f \in L^1(\mu)$. For $x \in [0, 1]$, the singleton $\{x\}$ is countable, so $\chi_{\{x\}} \in L^1(\mu)$, and $$x = \Lambda(\chi_{\{x\}}) = \int \chi_{\{x\}} h \, d\mu = h(x).$$ So $h = g$, which is not measurable. Contradiction. ($\mu$ is not σ-finite: a countable union of sets of finite counting measure is countable.) $\blacksquare$
So for $p = 1$, σ-finiteness is exactly what keeps the representing function measurable and unique. Next, the other half of the title.
From here on, $X$ is a locally compact Hausdorff space, and the space is $C_0(X)$, the continuous functions that vanish at infinity, with the sup norm. Its natural functionals are integrals against complex measures. To write those down we need the integral first.
On The Radon–Nikodym Theorem, every complex measure $\mu$ came with a measurable $h$, $|h| = 1$, such that $d\mu = h \, d|\mu|$.
Let $\mu$ be a complex measure with polar decomposition $d\mu = h \, d|\mu|$. For $f \in L^1(|\mu|)$, the integral of $f$ with respect to $\mu$ is $\displaystyle\int_X f \, d\mu = \int_X f h \, d|\mu|$.
It does what you'd hope, even when the masses are complex numbers. Here's the smallest interesting case.
Let $\mu = 2\delta_0 + (1 - i)\delta_1$ on the Borel sets of $\mathbb{R}$, and $f(x) = x^2 + 1$. Find $|\mu|$, the polar part $h$, and $\int f \, d\mu$.
An atom's total variation is the modulus of its mass, so $|\mu| = 2\delta_0 + \sqrt{2}\,\delta_1$, since $|1 - i| = \sqrt{2}$. Then $d\mu = h \, d|\mu|$ needs $h(0) \cdot 2 = 2$ and $h(1)\sqrt{2} = 1 - i$: so $h(0) = 1$ and $h(1) = \frac{1 - i}{\sqrt{2}}$, both of modulus $1$. (Off $\{0, 1\}$, $h$ can be anything of modulus $1$; $|\mu|$ doesn't see it.) Now $$\int f \, d\mu = \int f h \, d|\mu| = f(0) \cdot 1 \cdot 2 + f(1) \cdot \frac{1 - i}{\sqrt{2}} \cdot \sqrt{2} = 2 + 2(1 - i) = 4 - 2i.$$ That's $2f(0) + (1 - i)f(1)$, exactly what the masses say.
In general, for $\mu = \sum_k c_k \delta_{x_k}$ the same steps give $\int f \, d\mu = \sum_k c_k f(x_k)$. And since $|h| = 1$, always $\left|\int f \, d\mu\right| \le \int |f| \, d|\mu|$. The measures that match the topology of $X$ get a name.
A complex Borel measure $\mu$ on $X$ is regular if its total variation $|\mu|$ is regular. The set of regular complex Borel measures on $X$, with the norm $\|\mu\| = |\mu|(X)$, is written $M(X)$.
On $\mathbb{R}^n$ this is no restriction at all: every finite positive Borel measure there is regular, so $M(\mathbb{R}^n)$ is every complex Borel measure. On stranger spaces the word matters. Either way, $M(X)$ is a Banach space, which the last page almost proved.
Prove that $M(X)$ is a Banach space. (The Spicy Bonus on Total Variation showed that all complex measures on the Borel sets, with $\|\mu\| = |\mu|(X)$, form a Banach space; show that $M(X)$ is a closed subspace.)
Proof. For finite positive measures, regularity of $\rho$ says: for every Borel $E$ and $\varepsilon > 0$ there are a compact $K$ and an open $V$ with $K \subseteq E \subseteq V$ and $\rho(V \setminus K) < \varepsilon$. (Finite measures let us subtract.)
Domination. If $\rho$ is finite, positive, and regular, and $\nu$ is a positive measure with $\nu(A) \le \rho(A)$ for every Borel $A$, then $\nu$ is regular: the same $K$ and $V$ give $\nu(V \setminus K) \le \rho(V \setminus K) < \varepsilon$.
Subspace. If $\rho_1$, $\rho_2$ are finite and regular, so is $\rho_1 + \rho_2$: take $K_i \subseteq E \subseteq V_i$ for each with error $\frac{\varepsilon}{2}$, and use $K_1 \cup K_2$ and $V_1 \cap V_2$. Now let $\mu, \nu \in M(X)$ and $c \in \mathbb{C}$. For any partition $(E_n)$ of $A$, $\sum |\mu(E_n) + c\nu(E_n)| \le \sum |\mu(E_n)| + |c| \sum |\nu(E_n)|$, so $|\mu + c\nu| \le |\mu| + |c||\nu|$. By domination, $\mu + c\nu \in M(X)$.
Closed. Let $\mu_n \in M(X)$ with $|\mu - \mu_n|(X) \to 0$. Given $E$ and $\varepsilon$, pick $n$ with $|\mu - \mu_n|(X) < \varepsilon$, then $K \subseteq E \subseteq V$ with $|\mu_n|(V \setminus K) < \varepsilon$. As above, $|\mu| \le |\mu - \mu_n| + |\mu_n|$, so $|\mu|(V \setminus K) < 2\varepsilon$. So $\mu \in M(X)$. A closed subspace of a Banach space is a Banach space. $\blacksquare$
Now the functionals. Each $\mu \in M(X)$ gives $\Lambda_\mu(f) = \int_X f \, d\mu$ on $C_0(X)$, and its norm is the total mass of $|\mu|$, cancellations and all ignored.
Let $\mu \in M(X)$. Then $\Lambda_\mu(f) = \int_X f \, d\mu$ is a bounded linear functional on $C_0(X)$ with $\|\Lambda_\mu\| = |\mu|(X)$.
Proof. Let $d\mu = h \, d|\mu|$. Every $f \in C_0(X)$ is bounded and Borel, and $|\mu|$ is finite, so $f \in L^1(|\mu|)$ and $$|\Lambda_\mu(f)| = \left|\int_X fh \, d|\mu|\right| \le \int_X |f| \, d|\mu| \le \|f\|_\infty \, |\mu|(X).$$ So $\|\Lambda_\mu\| \le |\mu|(X)$.
For the reverse, the ideal test function is $\overline{h}$, since $\int \overline{h} \, d\mu = \int \overline{h}h \, d|\mu| = |\mu|(X)$. But $\overline{h}$ is only Borel, not continuous. Lusin's theorem fixes that. The finite regular measure $|\mu|$ satisfies the standing hypotheses on the Borel sets (it is outer regular, every set has finite measure and so is inner regular, and compact sets have finite measure), and $\overline{h}$ is nonzero only on $X$, which has finite measure. So Lusin's theorem gives $g_n \in C_c(X) \subseteq C_0(X)$ with $$|\mu|(\{g_n \ne \overline{h}\}) < \frac{1}{n} \quad \text{and} \quad \|g_n\|_\infty \le \sup|\overline{h}| = 1.$$ Then $$\big|\Lambda_\mu(g_n) - |\mu|(X)\big| = \left|\int_X (g_n - \overline{h}) h \, d|\mu|\right| \le \int_{\{g_n \ne \overline{h}\}} |g_n - \overline{h}| \, d|\mu| \le \frac{2}{n}.$$ So $\|\Lambda_\mu\| \ge |\Lambda_\mu(g_n)| \ge |\mu|(X) - \frac{2}{n}$ for every $n$. $\blacksquare$
So $\mu \mapsto \Lambda_\mu$ is an isometry from $M(X)$ into $C_0(X)^*$. Here's the approximation in the proof, for a measure with a positive half and a negative half: continuous ramps close in on the sign $\overline{h}$, and $\Lambda_\mu$ of them climbs to $|\mu|(X)$.
The norm counts every piece of $\mu$ with its own sign removed. That's easy to forget when the pieces are mixed together in one formula.
On $C_0(\mathbb{R})$, let $\Phi(f) = 2f(0) - 3f(1) + \int_0^\infty f(x) e^{-x} \, dx$. Find the measure $\mu$ with $\Phi = \Lambda_\mu$, and $\|\Phi\|$.
$\mu = 2\delta_0 - 3\delta_1 + e^{-x}\chi_{[0, \infty)} \, dm$. The atoms sit on $\{0, 1\}$, a Lebesgue-null set, and the density part lives off it, so the three pieces are concentrated on disjoint sets and the total variation adds up piece by piece: $$|\mu| = 2\delta_0 + 3\delta_1 + e^{-x}\chi_{[0, \infty)} \, dm, \qquad \|\Phi\| = |\mu|(\mathbb{R}) = 2 + 3 + \int_0^\infty e^{-x} \, dx = 6.$$ The near-maximizers are continuous $f$ with $|f| \le 1$ that are $1$ at $0$, $-1$ at $1$, and $1$ on most of $[0, \infty)$, switching sign in a tiny window around $1$. The value $|\mu(\mathbb{R})| = |2 - 3 + 1| = 0$ is the trap: cancellation is exactly what the norm ignores.
The Proposition put $M(X)$ inside $C_0(X)^*$. The Riesz representation theorem says it's all of it. We already have a Riesz theorem, for positive functionals.
On The Riesz Representation Theorem, every positive linear functional $\Lambda$ on $C_c(X)$ was $\Lambda f = \int f \, d\lambda$ for a positive measure $\lambda$, outer regular, inner regular on open sets and on sets of finite measure, and finite on compact sets.
A bounded functional $\Phi$ can be negative or complex, so that theorem doesn't apply directly. The plan: build a positive functional $\Lambda$ that dominates $\Phi$, get a measure from it, and then find $\Phi$'s density against that measure using the $L^1$ duality we just proved. The positive functional is built from an envelope: $\Lambda f$ is the best $|\Phi(h)|$ you can get from an $h$ that fits under $f$.
The hard part is showing that this sup is additive. Then it extends to a positive linear functional.
Let $\Phi$ be a linear functional on $C_c(X)$ with $|\Phi(f)| \le \|f\|_\infty$ for every $f$. Then there is a positive linear functional $\Lambda$ on $C_c(X)$ with $$|\Phi(f)| \le \Lambda(|f|) \le \|f\|_\infty \quad \text{for every } f \in C_c(X).$$
Proof. Step 1: the envelope. For $f \in C_c(X)$ with $f \ge 0$, let $$\Lambda f = \sup\{|\Phi(h)| : h \in C_c(X), \ |h| \le f\}.$$ Then $0 \le \Lambda f \le \|f\|_\infty$ (each such $h$ has $|\Phi(h)| \le \|h\|_\infty \le \|f\|_\infty$), $|\Phi(f)| \le \Lambda f$ (take $h = f$), $\Lambda f_1 \le \Lambda f_2$ when $f_1 \le f_2$, and $\Lambda(cf) = c\Lambda f$ for $c \ge 0$ (for $c > 0$, $|h| \le cf$ exactly when $|h/c| \le f$).
Step 2: $\Lambda f + \Lambda g \le \Lambda(f + g)$. Let $f, g \ge 0$ and $\varepsilon > 0$. Pick $h_1, h_2 \in C_c(X)$ with $|h_1| \le f$, $|h_2| \le g$, $\Lambda f \le |\Phi(h_1)| + \varepsilon$, and $\Lambda g \le |\Phi(h_2)| + \varepsilon$. Pick $\alpha_i \in \mathbb{C}$ with $|\alpha_i| = 1$ and $\alpha_i \Phi(h_i) = |\Phi(h_i)|$. Since $|\alpha_1 h_1 + \alpha_2 h_2| \le f + g$, $$\Lambda f + \Lambda g \le \Phi(\alpha_1 h_1 + \alpha_2 h_2) + 2\varepsilon \le |\Phi(\alpha_1 h_1 + \alpha_2 h_2)| + 2\varepsilon \le \Lambda(f + g) + 2\varepsilon.$$
Step 3: $\Lambda(f + g) \le \Lambda f + \Lambda g$. Let $h \in C_c(X)$ with $|h| \le f + g$, and let $V = \{f + g > 0\}$. Define $$h_1 = \frac{f h}{f + g}, \quad h_2 = \frac{g h}{f + g} \quad \text{on } V, \qquad h_1 = h_2 = 0 \text{ off } V.$$ Both are continuous on the open set $V$. At a point $x_0 \notin V$, $f(x_0) + g(x_0) = 0$, so $h(x_0) = 0$; since $|h_i| \le |h|$ everywhere and $h$ is continuous with $h(x_0) = 0$, $h_i$ is continuous at $x_0$. Both vanish off $\operatorname{supp} h$, so $h_1, h_2 \in C_c(X)$. Also $|h_1| \le f$, $|h_2| \le g$, and $h_1 + h_2 = h$. So $|\Phi(h)| \le |\Phi(h_1)| + |\Phi(h_2)| \le \Lambda f + \Lambda g$. Take the sup over $h$.
Step 4: extend. For real $f \in C_c(X)$, set $\Lambda f = \Lambda f^+ - \Lambda f^-$. If $f = u - v$ with $u, v \ge 0$ in $C_c(X)$, then $f^+ + v = f^- + u$, so additivity gives $\Lambda f^+ + \Lambda v = \Lambda f^- + \Lambda u$, that is, $\Lambda f = \Lambda u - \Lambda v$. Applied to $f + g = (f^+ + g^+) - (f^- + g^-)$, this makes $\Lambda$ additive on real functions; $\Lambda(cf) = c\Lambda f$ for $c \ge 0$ by Step 1, and $\Lambda(-f) = \Lambda f^- - \Lambda f^+ = -\Lambda f$. For complex $f = u + iv$, set $\Lambda f = \Lambda u + i\Lambda v$; this is complex-linear (for instance $\Lambda(if) = \Lambda(-v) + i\Lambda u = i\Lambda f$). It is positive by construction, and for every $f \in C_c(X)$, Step 1 applied to the envelope $|f|$ gives $|\Phi(f)| \le \Lambda(|f|) \le \|f\|_\infty$. $\blacksquare$
Step 3 is a split: any $h$ that fits under $f + g$ breaks into a piece under $f$ and a piece under $g$, in proportion. Try to find the split yourself before you look at the formula's answer.
Now the theorem. Two more facts from earlier pages do the gluing: $C_c(X)$ is dense in $C_0(X)$, and we'll use the $p = 1$ case of the duality theorem against the measure the lemma produces.
On Inclusions & Approximation, $C_0(X)$ turned out to be the completion of $C_c(X)$ in the sup norm: it is a Banach space, and $C_c(X)$ is dense in it.
Let $X$ be a locally compact Hausdorff space. Every bounded linear functional $\Phi$ on $C_0(X)$ is represented by a unique regular complex Borel measure $\mu$: $$\Phi(f) = \int_X f \, d\mu \quad \text{for every } f \in C_0(X).$$ Moreover, $\|\Phi\| = |\mu|(X)$. So $\mu \mapsto \Lambda_\mu$ is an isometric isomorphism of $M(X)$ onto $C_0(X)^*$: $C_0(X)^* \cong M(X)$ isometrically.
Proof. Uniqueness and the norm. If $\mu$ and $\nu$ in $M(X)$ both represent $\Phi$, then $\mu - \nu \in M(X)$ (the Exercise above) and $\Lambda_{\mu - \nu} = 0$, so the Proposition gives $|\mu - \nu|(X) = \|\Lambda_{\mu - \nu}\| = 0$. Once $\mu$ exists, $\|\Phi\| = \|\Lambda_\mu\| = |\mu|(X)$. So it remains to find $\mu$. If $\Phi = 0$, take $\mu = 0$; otherwise divide by $\|\Phi\|$ and assume $\|\Phi\| = 1$.
Step 1: a dominating measure. The lemma, applied to $\Phi$ on $C_c(X)$, gives a positive linear $\Lambda$ on $C_c(X)$ with $|\Phi(f)| \le \Lambda(|f|) \le \|f\|_\infty$. The positive Riesz theorem gives a positive measure $\lambda$, on a σ-algebra containing the Borel sets, with $\Lambda f = \int_X f \, d\lambda$ for $f \in C_c(X)$ and the regularity properties listed there. Restrict $\lambda$ to the Borel sets.
Step 2: $\lambda(X) \le 1$, so $\lambda$ is regular. $X$ is open, so $\lambda(X) = \sup\{\lambda(K) : K \subseteq X \text{ compact}\}$. For compact $K$, Urysohn's lemma gives $f \in C_c(X)$ with $\chi_K \le f \le 1$, so $\lambda(K) \le \int_X f \, d\lambda = \Lambda f \le \|f\|_\infty = 1$. So $\lambda(X) \le 1$. Every Borel set then has finite measure, so it is inner regular as well as outer regular: $\lambda$ is a finite regular Borel measure.
Step 3: a density. For $f \in C_c(X)$, $$|\Phi(f)| \le \Lambda(|f|) = \int_X |f| \, d\lambda = \|f\|_{L^1(\lambda)}.$$ So $\Phi$ is a bounded functional on $C_c(X)$ in the $L^1(\lambda)$ norm, with norm at most $1$ (functions equal $\lambda$-a.e. get the same value, since $\Phi$ of their difference is at most its $L^1$ norm, $0$). $C_c(X)$ is dense in $L^1(\lambda)$ ($\lambda$ satisfies the standing hypotheses), so $\Phi$ extends by continuity to a bounded functional on $L^1(\lambda)$, still with norm at most $1$. Since $\lambda$ is finite, the duality theorem with $p = 1$ gives $g \in L^\infty(\lambda)$ with $\|g\|_\infty \le 1$ and $$\Phi(f) = \int_X fg \, d\lambda \quad \text{for every } f \in C_c(X).$$
Step 4: all of $C_0(X)$. Both sides are continuous on $C_0(X)$ in the sup norm: $\Phi$ by assumption, and $\left|\int_X fg \, d\lambda\right| \le \|f\|_\infty \|g\|_\infty \lambda(X)$. They agree on $C_c(X)$, which is dense in $C_0(X)$, so they agree on $C_0(X)$.
Step 5: the measure. Let $\mu(E) = \int_E g \, d\lambda$ for Borel $E$. Since $g$ is bounded and $\lambda$ is finite, $g \in L^1(\lambda)$, so $\mu$ is a complex measure, and by the corollary of the polar decomposition, $|\mu|(E) = \int_E |g| \, d\lambda$. Write $d\mu = h \, d|\mu|$. Then $\int_E g \, d\lambda = \mu(E) = \int_E h \, d|\mu| = \int_E h|g| \, d\lambda$ for every $E$, so $g = h|g|$ $\lambda$-a.e., and for $f \in C_0(X)$, $$\int_X f \, d\mu = \int_X fh \, d|\mu| = \int_X fh|g| \, d\lambda = \int_X fg \, d\lambda = \Phi(f).$$ Finally, $\mu$ is regular: $|\mu|(E) = \int_E |g| \, d\lambda \le \lambda(E)$, and $\lambda$ is finite and regular, so the domination step of the Exercise above applies. $\blacksquare$
As a bonus, $1 = \|\Phi\| = |\mu|(X) = \int_X |g| \, d\lambda \le \lambda(X) \le 1$ forces $\lambda(X) = 1$ and $|g| = 1$ $\lambda$-a.e. So the measure the lemma built was $|\mu|$ all along.
So $C_0(X)^*$ is $M(X)$, measures and all. That's the real difference from $L^p$: a functional on $C_0$ can concentrate on a single point, like $f \mapsto f(0)$, and no function $g$ can do that. The widget below hides a measure behind a functional. Probe it with tents and hunt down what's behind it.
One more worked example, with infinitely many atoms. The norm is still just the total of the absolute masses.
On $C_0(\mathbb{R})$, let $\Phi(f) = \sum_{n=1}^\infty (-1)^n 2^{-n} f(n) + \int_{-1}^{1} x f(x) \, dx$. Find the representing measure and $\|\Phi\|$.
The series converges absolutely, and $|\Phi(f)| \le \|f\|_\infty \left(\sum 2^{-n} + \int_{-1}^1 |x| \, dx\right)$, so $\Phi$ is bounded. Its measure is $$\mu = \sum_{n=1}^\infty (-1)^n 2^{-n} \delta_n + x\chi_{[-1, 1]} \, dm.$$ The atoms live on $\mathbb{N}$, a Lebesgue-null set, so the two parts are concentrated on disjoint sets and $|\mu| = \sum 2^{-n}\delta_n + |x|\chi_{[-1, 1]} \, dm$. So $$\|\Phi\| = |\mu|(\mathbb{R}) = \sum_{n=1}^\infty 2^{-n} + \int_{-1}^{1} |x| \, dx = 1 + 1 = 2.$$
Notice that the atom at $n = 1$ sits inside $[-1, 1]$, where the density lives. That doesn't matter: a single point carries no Lebesgue mass, so the two parts still don't overlap.
The sup-norm space matters. $C_c(X)$ with the sup norm has the same dual, $M(X)$, because it is dense in $C_0(X)$ and a bounded functional extends uniquely from a dense subspace. But $C_b(\mathbb{R})$, all bounded continuous functions, has a bigger dual: extend $f \mapsto \lim_{x \to \infty} f(x)$ from the functions that have a limit, by Hahn–Banach. That functional kills all of $C_0(\mathbb{R})$, so its measure would have to be $0$, yet it sends the constant $1$ to $1$.
Last lesson question: name the dual. Decide which theorem applies, then read off the answer.
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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 proofs. Each one is a line from a proof on this page with a gap in it. Fill the gap.
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Then 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 that need their answer choices will say so.
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Last, the Spicy Bonus. It's a converse to Hölder: if $fg$ is integrable for every $f \in L^p$, then $g$ must already be in $L^q$. The tempting proof says "each $\int fg$ is finite, so $f \mapsto \int fg$ is bounded." Finite at every point is not the same as bounded, and closing that gap takes a theorem from a few pages back.
On Baire Category: a family of bounded linear maps on a Banach space that is bounded at every point is bounded in norm.
Let $1 \le p \le \infty$, let $q$ be the conjugate exponent, let $\mu$ be a σ-finite positive measure, and let $g$ be a measurable complex function such that $fg \in L^1(\mu)$ for every $f \in L^p(\mu)$. Prove that $g \in L^q(\mu)$.
Proof. Truncate. Write $X = \bigcup_n X_n$ with $X_1 \subseteq X_2 \subseteq \cdots$ and $\mu(X_n) < \infty$ (finite unions of a σ-finite cover). Let $G_n = X_n \cap \{|g| \le n\}$ and $g_n = g\chi_{G_n}$. Then $|g_n| \le n$ and $g_n$ vanishes off a set of finite measure, so $g_n \in L^q(\mu)$, for every $q$.
A family of functionals. Let $\Lambda_n = \Lambda_{g_n}$ on $L^p(\mu)$. By the Proposition (with σ-finiteness covering $p = 1$), each $\Lambda_n$ is bounded with $\|\Lambda_n\| = \|g_n\|_q$. For each fixed $f \in L^p(\mu)$, $$|\Lambda_n(f)| \le \int_X |f g_n| \, d\mu \le \int_X |fg| \, d\mu < \infty,$$ by the hypothesis, and the bound doesn't depend on $n$. So $\sup_n |\Lambda_n(f)| < \infty$ for every $f$.
Uniform boundedness. $L^p(\mu)$ is a Banach space for every $1 \le p \le \infty$ (The Lᵖ Spaces), so the uniform boundedness principle gives $M = \sup_n \|\Lambda_n\| < \infty$. That is, $\|g_n\|_q \le M$ for every $n$.
Pass to the limit. The sets $G_n$ increase to $X$: $X_n \uparrow X$, and $\{|g| \le n\} \uparrow X$ because $g$ is complex-valued. If $q < \infty$, then $|g_n|^q \uparrow |g|^q$, and monotone convergence gives $\int_X |g|^q \, d\mu = \lim_n \|g_n\|_q^q \le M^q$. If $q = \infty$, then $|g| \le M$ a.e. on each $G_n$, so off the null set $\bigcup_n (G_n \cap \{|g| > M\})$, which means $|g| \le M$ a.e. Either way, $g \in L^q(\mu)$. $\blacksquare$