[proofplan] The map $\varphi: g \mapsto \varphi_g$ has $\|\varphi_g\| \le \|g\|_q$ from Hölder's inequality (already noted in the definition); we improve this to equality by exhibiting test functions $f$ where $|\varphi_g(f)| / \|f\|_p$ approaches $\|g\|_q$. For $p \in (1, \infty)$ the test function $f = \lambda |g|^{q-1}$ saturates Hölder; for $p = 1$ a normalised indicator $f = \lambda \mathbb{1}_B$ on a level set of $|g|$ approaches $\|g\|_\infty$. For surjectivity, given $\psi \in L^p(\mu)^*$ we first treat $\mu$ finite: the set function $\nu(A) := \psi(\mathbb{1}_A)$ is a complex measure with $\nu \ll \mu$, and the [Radon-Nikodym Theorem](/theorems/2640) yields a density $g \in L^1(\mu)$. We then bound $g$ in $L^q$ by testing $\psi$ against truncated witnesses, and extend by density of $L^\infty$ in $L^p$. The $\sigma$-finite case follows by partitioning, and the general case for $p > 1$ follows by localising to a $\sigma$-finite subset where $\|\psi\|$ is realised. [/proofplan] [step:Recall the upper bound $\|\varphi_g\| \le \|g\|_q$ from Hölder] Fix $g \in L^q(\mu)$. For $f \in L^p(\mu)$, Hölder's inequality (with exponents $p$ and $q$ satisfying $1/p + 1/q = 1$) gives \begin{align*} |\varphi_g(f)| = \left|\int_\Omega fg \, d\mu\right| \le \int_\Omega |fg| \, d\mu \le \|f\|_p \|g\|_q. \end{align*} Therefore $\varphi_g \in L^p(\mu)^*$ with $\|\varphi_g\| \le \|g\|_q$. The map $\varphi: g \mapsto \varphi_g$ is linear (immediate from linearity of the integral). To show $\varphi$ is an isometry, it suffices to prove $\|\varphi_g\| \ge \|g\|_q$ for all $g \in L^q(\mu)$ in both cases. [/step] [step:Construct a polar function $\lambda$ that rotates $g$ to non-negative real values] For both cases we use the following auxiliary function. Define \begin{align*} \lambda: \Omega &\to \{z \in \mathbb{C} : |z| = 1\} \cup \{0\} \\ \omega &\mapsto \begin{cases} \overline{g(\omega)}/|g(\omega)| & g(\omega) \ne 0, \\ 0 & g(\omega) = 0. \end{cases} \end{align*} Then $\lambda$ is measurable (composition of measurable functions, with the convention $0 / 0 = 0$ on the null set $\{g = 0\}$ where its value does not matter for the integrals) and $|\lambda| \le 1$ everywhere, with $\lambda g = |g|$ a.e. (on $\{g \ne 0\}$ by construction, on $\{g = 0\}$ both sides are $0$). [/step] [step:Prove the lower bound $\|\varphi_g\| \ge \|g\|_q$ for $p \in (1, \infty)$] Assume $p \in (1, \infty)$, so $q \in (1, \infty)$. We may assume $g \ne 0$ in $L^q$, else there is nothing to prove. Define \begin{align*} f: \Omega &\to \mathbb{C} \\ \omega &\mapsto \lambda(\omega) |g(\omega)|^{q-1}. \end{align*} We compute $\|f\|_p$. Using $|\lambda| \le 1$ with equality where $g \ne 0$, $|f|^p = |g|^{p(q-1)}$ on $\{g \ne 0\}$ and is $0$ elsewhere; the relation $1/p + 1/q = 1$ rearranges to $p(q-1) = q$, so \begin{align*} \|f\|_p^p = \int_\Omega |f|^p \, d\mu = \int_\Omega |g|^q \, d\mu = \|g\|_q^q, \end{align*} giving $\|f\|_p = \|g\|_q^{q/p} \in (0, \infty)$, so $f \in L^p(\mu)$. Next, $\varphi_g(f) = \int_\Omega \lambda |g|^{q-1} g \, d\mu = \int_\Omega |g| \cdot |g|^{q-1} \, d\mu = \int_\Omega |g|^q \, d\mu = \|g\|_q^q$, where we used $\lambda g = |g|$ a.e. So \begin{align*} \|\varphi_g\| \ge \frac{|\varphi_g(f)|}{\|f\|_p} = \frac{\|g\|_q^q}{\|g\|_q^{q/p}} = \|g\|_q^{q - q/p} = \|g\|_q^{q(1 - 1/p)} = \|g\|_q^{q/q} = \|g\|_q, \end{align*} where the last step uses $1 - 1/p = 1/q$. Combined with the upper bound from Step 1, $\|\varphi_g\| = \|g\|_q$. [guided] The choice $f = \lambda |g|^{q-1}$ is dictated by saturating Hölder's inequality. Recall: Hölder gives $|\int fg| \le (\int |f|^p)^{1/p} (\int |g|^q)^{1/q}$, with equality iff $|f|^p$ and $|g|^q$ are proportional and $\arg(fg)$ is a.e. constant. Setting $|f|^p = |g|^q$ (proportional with constant 1) gives $|f| = |g|^{q/p} = |g|^{q-1}$ (using $q/p = q - 1$, which is $1/p + 1/q = 1$ rearranged). Multiplying by $\lambda$ aligns the phase: $\lambda g = |g|$ a.e., so $fg = \lambda |g|^{q-1} g = |g|^q$, which is non-negative and real, the equality case of Hölder. Computing both sides of Hölder for this choice gives $\int |g|^q$ on the LHS and $(\int |g|^q)^{1/p + 1/q} = (\int |g|^q)^1 = \int |g|^q$ on the RHS — they really are equal. The arithmetic of exponents $q - q/p = q/q = 1$ then converts this to $\|\varphi_g\| \ge \|g\|_q$. The hypothesis $p \in (1, \infty)$ is used because $|g|^{q-1}$ requires $q - 1 \ge 0$, i.e. $q \ge 1$ (true), but more importantly to make $|g|^{q-1}$ lie in $L^p$ via the exponent identity $p(q-1) = q$. For $p = 1$ we have $q = \infty$ and the test function would be $|g|^{\infty - 1}$, which is meaningless; we need a different witness, supplied below. [/guided] [/step] [step:Prove the lower bound $\|\varphi_g\| \ge \|g\|_\infty$ for $p = 1$ (using $\sigma$-finiteness)] Assume $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite. We may assume $\|g\|_\infty > 0$. Fix any $s \in (0, \|g\|_\infty)$. By definition of $\|g\|_\infty = \operatorname{ess\,sup}|g|$, the set \begin{align*} A := \{\omega \in \Omega : |g(\omega)| > s\} \end{align*} satisfies $\mu(A) > 0$. By $\sigma$-finiteness, write $\Omega = \bigsqcup_{n=1}^\infty C_n$ with $\mu(C_n) < \infty$. Then $A = \bigsqcup_n (A \cap C_n)$, so $\sum_n \mu(A \cap C_n) = \mu(A) > 0$, and at least one $A \cap C_n$ has positive measure. Let $B := A \cap C_n$ for some such $n$; then $B \in \Sigma$, $B \subseteq A$, and $0 < \mu(B) \le \mu(C_n) < \infty$. Define $f := \lambda \mathbb{1}_B$. Then $|f| = |\lambda| \mathbb{1}_B \le \mathbb{1}_B$, so $\|f\|_1 = \int_B |\lambda| \, d\mu \le \mu(B) < \infty$, and in fact $\|f\|_1 = \int_B |\lambda| \, d\mu = \mu(B \cap \{g \ne 0\}) = \mu(B)$ since $B \subseteq A \subseteq \{|g| > s > 0\} \subseteq \{g \ne 0\}$. So $\|f\|_1 = \mu(B)$. Compute \begin{align*} \varphi_g(f) = \int_\Omega \lambda \mathbb{1}_B g \, d\mu = \int_B \lambda g \, d\mu = \int_B |g| \, d\mu \ge \int_B s \, d\mu = s \mu(B), \end{align*} using $\lambda g = |g|$ a.e. and $|g| > s$ on $B$. Therefore \begin{align*} \|\varphi_g\| \ge \frac{|\varphi_g(f)|}{\|f\|_1} \ge \frac{s \mu(B)}{\mu(B)} = s. \end{align*} This holds for every $s \in (0, \|g\|_\infty)$, so $\|\varphi_g\| \ge \|g\|_\infty$. Combined with Step 1, $\|\varphi_g\| = \|g\|_\infty$. [guided] Why does the case $p = 1$ require $\sigma$-finiteness? The lower bound $\|\varphi_g\| \ge s$ comes from a test function $f = \lambda \mathbb{1}_B$ on a set $B$ where $|g| > s$. This $f$ must lie in $L^1$, requiring $\mu(B) < \infty$. So we need a measurable set $B \subseteq \{|g| > s\}$ of strictly positive but finite measure. Without $\sigma$-finiteness this can fail: there could be a set $A = \{|g| > s\}$ of positive but only "totally infinite" measure — every measurable subset is either null or has infinite measure. (This is what happens for non-$\sigma$-finite atoms.) In that pathological case, $\varphi$ would not be an isometry. With $\sigma$-finiteness, the partition $\Omega = \bigsqcup_n C_n$ slices any positive-measure set into finite pieces; some piece $A \cap C_n$ has positive finite measure, giving us our $B$. The factor $\mu(B)$ cancels exactly between numerator $s\mu(B)$ and denominator $\|f\|_1 = \mu(B)$, so the actual size of $\mu(B)$ does not matter — only that it is positive and finite. We get $\|\varphi_g\| \ge s$, and taking $s \uparrow \|g\|_\infty$ gives $\|\varphi_g\| \ge \|g\|_\infty$. [/guided] [/step] [step:Prove surjectivity for finite $\mu$, $p \in [1, \infty)$ via Radon-Nikodym] Now we turn to surjectivity. Assume $\mu$ is a finite positive measure (so $\mu(\Omega) < \infty$) and fix $\psi \in L^p(\mu)^*$ with $p \in [1, \infty)$. We construct $g \in L^q(\mu)$ with $\psi = \varphi_g$. For $A \in \Sigma$, $\mathbb{1}_A \in L^p(\mu)$ since $\|\mathbb{1}_A\|_p = \mu(A)^{1/p} \le \mu(\Omega)^{1/p} < \infty$. Define \begin{align*} \nu: \Sigma &\to \mathbb{C} \\ A &\mapsto \psi(\mathbb{1}_A). \end{align*} **$\nu$ is finitely additive.** If $A, B \in \Sigma$ are disjoint, $\mathbb{1}_{A \cup B} = \mathbb{1}_A + \mathbb{1}_B$, so $\nu(A \cup B) = \psi(\mathbb{1}_A + \mathbb{1}_B) = \psi(\mathbb{1}_A) + \psi(\mathbb{1}_B) = \nu(A) + \nu(B)$. **$\nu$ is countably additive.** If $A_k \in \Sigma$ are pairwise disjoint with $A := \bigsqcup_k A_k$, set $S_n := \sum_{k=1}^n \mathbb{1}_{A_k} = \mathbb{1}_{\bigsqcup_{k \le n} A_k}$. Then $|S_n - \mathbb{1}_A|^p = \mathbb{1}_{A \setminus \bigsqcup_{k \le n} A_k}$, so $\|S_n - \mathbb{1}_A\|_p^p = \mu(A \setminus \bigsqcup_{k \le n} A_k) \to 0$ by countable additivity of $\mu$ (and finiteness $\mu(A) < \infty$). So $S_n \to \mathbb{1}_A$ in $L^p(\mu)$, and continuity of $\psi$ gives $\sum_{k=1}^n \nu(A_k) = \psi(S_n) \to \psi(\mathbb{1}_A) = \nu(A)$. **$\nu \ll \mu$.** If $\mu(A) = 0$ then $\mathbb{1}_A = 0$ in $L^p(\mu)$, so $\nu(A) = \psi(0) = 0$. By the [Radon-Nikodym Theorem](/theorems/2640) applied to the $\sigma$-finite (in fact finite) measure space $(\Omega, \Sigma, \mu)$ and the complex measure $\nu \ll \mu$, there exists a unique $g \in L^1(\mu)$ such that $\nu(A) = \int_A g \, d\mu$ for all $A \in \Sigma$. So \begin{align*} \psi(\mathbb{1}_A) = \int_A g \, d\mu = \int_\Omega \mathbb{1}_A g \, d\mu = \varphi_g(\mathbb{1}_A). \end{align*} [guided] Why does the [Radon-Nikodym Theorem](/theorems/2640) apply? Its hypotheses are: a $\sigma$-finite measure $\mu$ and a complex measure $\nu \ll \mu$. We verify both: - $\sigma$-finiteness of $\mu$: $\mu$ is finite, hence $\sigma$-finite (take $C_1 := \Omega$). - $\nu$ is a complex measure: $\nu(A) = \psi(\mathbb{1}_A)$ takes complex values; we showed countable additivity above (using continuity of $\psi$ in $L^p$ and $L^p$-convergence of indicator partial sums). - $\nu \ll \mu$: $\mu(A) = 0$ forces $\mathbb{1}_A = 0$ a.e., so $\mathbb{1}_A = 0$ in $L^p(\mu)$ (which is the equivalence-class space), so $\psi(\mathbb{1}_A) = 0$. The conclusion is a function $g \in L^1(\mu)$ with $\nu(A) = \int_A g \, d\mu$ for all $A$. Equivalently, $\psi(\mathbb{1}_A) = \int_\Omega \mathbb{1}_A \cdot g \, d\mu = \varphi_g(\mathbb{1}_A)$. So $\psi$ and $\varphi_g$ agree on indicator functions. Linearity then extends this agreement to simple functions. Note that at this stage we only know $g \in L^1$; we have not yet shown $g \in L^q$ — that requires further work in the next step. The point of having $g \in L^1$ is enough to define $\varphi_g(f) = \int fg \, d\mu$ for bounded $f$, and we use this to test against more functions and extract a sharper bound on $g$. [/guided] [/step] [step:Extend $\psi = \varphi_g$ from indicators to $L^\infty(\mu)$ by density] By linearity of both $\psi$ and $\varphi_g$ in $f$, the agreement $\psi(\mathbb{1}_A) = \varphi_g(\mathbb{1}_A)$ extends to all simple functions: for $s = \sum_{k=1}^N c_k \mathbb{1}_{A_k}$ with disjoint $A_k$ and $c_k \in \mathbb{C}$, \begin{align*} \psi(s) = \sum_k c_k \psi(\mathbb{1}_{A_k}) = \sum_k c_k \int_\Omega \mathbb{1}_{A_k} g \, d\mu = \int_\Omega s g \, d\mu = \varphi_g(s). \end{align*} Now extend to $L^\infty(\mu) \subset L^p(\mu)$ (the inclusion is valid because $\mu(\Omega) < \infty$: $\|f\|_p \le \mu(\Omega)^{1/p} \|f\|_\infty$). Fix $f \in L^\infty(\mu)$ with $\|f\|_\infty \le M$. There exists a sequence of simple functions $s_n$ with $|s_n| \le M$ and $s_n \to f$ pointwise $\mu$-a.e. (a standard approximation: split real/imaginary parts and use the canonical dyadic approximation of bounded measurable functions by simple ones). - $s_n \to f$ in $L^p(\mu)$: $|s_n - f|^p \le (2M)^p$ pointwise and $|s_n - f|^p \to 0$ a.e.; the dominated convergence theorem (dominator $(2M)^p \mathbb{1}_\Omega \in L^1(\mu)$ since $\mu(\Omega) < \infty$) gives $\|s_n - f\|_p^p \to 0$. - $\int_\Omega s_n g \, d\mu \to \int_\Omega f g \, d\mu$: $|s_n g| \le M|g|$ pointwise and $s_n g \to fg$ a.e.; the dominated convergence theorem (dominator $M|g| \in L^1(\mu)$ since $g \in L^1(\mu)$) gives the conclusion. Therefore $\psi(f) = \lim_n \psi(s_n) = \lim_n \int_\Omega s_n g \, d\mu = \int_\Omega fg \, d\mu = \varphi_g(f)$ for all $f \in L^\infty(\mu)$. [/step] [step:Bound $\|g\|_q$ for $p \in (1, \infty)$ by testing against truncated witnesses] We now upgrade $g \in L^1(\mu)$ to $g \in L^q(\mu)$ with $\|g\|_q \le \|\psi\|$. Assume $p \in (1, \infty)$ first. For $n \in \mathbb{N}$, define \begin{align*} B_n := \{|g| \le n\}, \qquad f_n := \lambda |g|^{q-1} \mathbb{1}_{B_n}, \end{align*} where $\lambda$ is from Step 2. On $B_n$, $|f_n| = |g|^{q-1} \le n^{q-1}$, so $f_n \in L^\infty(\mu) \subseteq L^p(\mu)$ (using $\mu(\Omega) < \infty$). Apply the identity from Step 7: \begin{align*} \psi(f_n) = \int_\Omega f_n g \, d\mu = \int_{B_n} \lambda |g|^{q-1} g \, d\mu = \int_{B_n} |g|^q \, d\mu, \end{align*} using $\lambda g = |g|$ a.e. Bound by the operator norm: \begin{align*} \int_{B_n} |g|^q \, d\mu = \psi(f_n) \le \|\psi\| \|f_n\|_p = \|\psi\| \left(\int_{B_n} |g|^{p(q-1)} \, d\mu\right)^{1/p} = \|\psi\| \left(\int_{B_n} |g|^q \, d\mu\right)^{1/p}, \end{align*} again using $p(q-1) = q$. Set $I_n := \int_{B_n} |g|^q \, d\mu \in [0, \infty)$ (finite because $|g|^q$ is bounded on $B_n$ and $\mu(\Omega) < \infty$). Then $I_n \le \|\psi\| I_n^{1/p}$, which on dividing by $I_n^{1/p}$ (valid if $I_n > 0$; the inequality holds with both sides zero when $I_n = 0$) gives \begin{align*} I_n^{1 - 1/p} = I_n^{1/q} \le \|\psi\|, \qquad I_n \le \|\psi\|^q. \end{align*} Now $|g|^q \mathbb{1}_{B_n} \uparrow |g|^q$ pointwise as $n \to \infty$ (since $\bigcup_n B_n \supseteq \{|g| < \infty\}$ and $|g| < \infty$ a.e. as $g \in L^1$). Monotone convergence gives \begin{align*} \int_\Omega |g|^q \, d\mu = \lim_n I_n \le \|\psi\|^q, \end{align*} so $g \in L^q(\mu)$ with $\|g\|_q \le \|\psi\|$. [/step] [step:Bound $\|g\|_\infty$ for $p = 1$] Now consider $p = 1$. Set $E := \{|g| > \|\psi\|\}$. We show $\mu(E) = 0$, i.e. $|g| \le \|\psi\|$ a.e., so $\|g\|_\infty \le \|\psi\|$. Suppose for contradiction $\mu(E) > 0$. Since $\mu$ is finite, $0 < \mu(E) < \infty$. Take $f := \lambda \mathbb{1}_E \in L^\infty(\mu) \subseteq L^1(\mu)$ (with $\|f\|_1 = \mu(E)$ as in Step 4). By Step 7, \begin{align*} \psi(f) = \int_\Omega \lambda \mathbb{1}_E g \, d\mu = \int_E |g| \, d\mu > \|\psi\| \mu(E) = \|\psi\| \|f\|_1, \end{align*} using $|g| > \|\psi\|$ on $E$. This contradicts $|\psi(f)| \le \|\psi\| \|f\|_1$. Hence $\mu(E) = 0$ and $\|g\|_\infty \le \|\psi\|$. [/step] [step:Extend $\psi = \varphi_g$ from $L^\infty$ to all of $L^p$ by density] We now have $g \in L^q(\mu)$ in both cases, so $\varphi_g \in L^p(\mu)^*$ is a well-defined bounded functional. We show $\psi = \varphi_g$ as elements of $L^p(\mu)^*$. We have $\psi(f) = \varphi_g(f)$ for all $f \in L^\infty(\mu)$ from Step 7. Since $\mu$ is finite, $L^\infty(\mu) \subseteq L^p(\mu)$ and $L^\infty(\mu)$ is dense in $L^p(\mu)$ (every $f \in L^p(\mu)$ is a limit in $L^p$ of truncations $f_n := f \mathbb{1}_{\{|f| \le n\}} \in L^\infty(\mu)$, by dominated convergence with dominator $|f|^p \in L^1(\mu)$). Both $\psi$ and $\varphi_g$ are continuous on $L^p(\mu)$, so they agree on the closure of $L^\infty(\mu)$, which is $L^p(\mu)$. Hence $\psi = \varphi_g$ in $L^p(\mu)^*$. This proves surjectivity in the case of finite $\mu$. [guided] The standard density argument: two continuous functions on a metric space that agree on a dense subset agree everywhere. Here: - The "metric space" is $L^p(\mu)$ with its norm topology. - The "dense subset" is $L^\infty(\mu) \subseteq L^p(\mu)$. - The "two continuous functions" are $\psi$ and $\varphi_g$, both bounded linear functionals on $L^p(\mu)$, hence continuous. - They agree on $L^\infty(\mu)$ from Step 7. Why is $L^\infty(\mu)$ dense in $L^p(\mu)$ when $\mu$ is finite? Given $f \in L^p(\mu)$, the truncations $f_n := f \mathbb{1}_{\{|f| \le n\}}$ are in $L^\infty(\mu)$ ($|f_n| \le n$). They converge to $f$ pointwise (since $|f| < \infty$ a.e., as $f \in L^p$ with $p < \infty$ and $\mu$ finite — actually $|f|^p < \infty$ a.e. is the fact, equivalent for $p < \infty$). The bound $|f_n - f|^p \le |f|^p$ provides an integrable dominator, and DCT gives $\|f_n - f\|_p \to 0$. [/guided] [/step] [step:Surjectivity for $\sigma$-finite $\mu$, $p \in [1, \infty)$] Now suppose $\mu$ is $\sigma$-finite. Partition $\Omega = \bigsqcup_n A_n$ with $\mu(A_n) < \infty$. For each $n$, restrict $\mu$ to $\Sigma_n := \{A \cap A_n : A \in \Sigma\}$ to get a finite measure space $(A_n, \Sigma_n, \mu|_{A_n})$. The restriction map $L^p(\mu|_{A_n}) \hookrightarrow L^p(\mu)$, $h \mapsto h \mathbb{1}_{A_n}$ (extending by zero outside $A_n$), is an isometric embedding. Define $\psi_n \in L^p(\mu|_{A_n})^*$ by $\psi_n(h) := \psi(h \mathbb{1}_{A_n})$ for $h \in L^p(\mu|_{A_n})$, with $\|\psi_n\| \le \|\psi\|$. By Step 10 applied to the finite measure $\mu|_{A_n}$, there exists $g_n \in L^q(\mu|_{A_n})$ with $\psi_n = \varphi_{g_n}$ and $\|g_n\|_{L^q(\mu|_{A_n})} \le \|\psi_n\| \le \|\psi\|$. Define $g: \Omega \to \mathbb{C}$ by $g(\omega) := g_n(\omega)$ for $\omega \in A_n$. This is well-defined since the $A_n$ partition $\Omega$, and measurable. We claim $g \in L^q(\mu)$ with $\|g\|_q \le \|\psi\|$: - For $q < \infty$: $\int_\Omega |g|^q \, d\mu = \sum_n \int_{A_n} |g_n|^q \, d\mu_n$. We need to show this sum is bounded by $\|\psi\|^q$. Test $\psi$ against the function $f := \lambda |g|^{q-1} \mathbb{1}_{B_N}$ where $B_N := \bigcup_{n=1}^N A_n \cap \{|g| \le N\}$ (which has finite measure and $|g|$ bounded, so $f \in L^\infty \subseteq L^p$). The same argument as in Step 8 gives $\int_{B_N} |g|^q \, d\mu \le \|\psi\|^q$. Letting $N \to \infty$, monotone convergence gives $\int_\Omega |g|^q \, d\mu \le \|\psi\|^q$. - For $q = \infty$: each $\|g_n\|_{\infty, A_n} \le \|\psi\|$, so $|g_n| \le \|\psi\|$ a.e. on $A_n$, and patching gives $|g| \le \|\psi\|$ a.e. on $\Omega$, so $\|g\|_\infty \le \|\psi\|$. For $A \in \Sigma$, by countable additivity (verified as in Step 6 using $\sum_n \|\mathbb{1}_{A \cap A_n}\|_p^p = \sum_n \mu(A \cap A_n) = \mu(A)$), \begin{align*} \psi(\mathbb{1}_A) = \sum_n \psi(\mathbb{1}_{A \cap A_n}) = \sum_n \psi_n(\mathbb{1}_{A \cap A_n}|_{A_n}) = \sum_n \int_{A \cap A_n} g_n \, d\mu_n = \int_A g \, d\mu = \varphi_g(\mathbb{1}_A). \end{align*} Density (as in Step 10) extends this to $\psi = \varphi_g$ on $L^p(\mu)$. [/step] [step:Surjectivity for general $\mu$, $p \in (1, \infty)$, by localising to a $\sigma$-finite carrier] Finally, let $\mu$ be an arbitrary measure (not necessarily $\sigma$-finite) and $p \in (1, \infty)$. Fix $\psi \in L^p(\mu)^*$. We claim there exists a $\sigma$-finite measurable set $\Omega_0 \subseteq \Omega$ such that $\psi(f) = \psi(f \mathbb{1}_{\Omega_0})$ for all $f \in L^p(\mu)$. Construct $\Omega_0$ as follows: let $\mathcal{S}$ denote the set of $\sigma$-finite measurable subsets of $\Omega$, and define \begin{align*} M := \sup\{\|\psi|_{L^p(S)}\| : S \in \mathcal{S}\}. \end{align*} Here $L^p(S) := \{f \in L^p(\mu) : f = 0 \text{ a.e. outside } S\}$, a closed subspace of $L^p(\mu)$. We have $M \le \|\psi\|$. Choose $S_n \in \mathcal{S}$ with $\|\psi|_{L^p(S_n)}\| > M - 1/n$, and set $\Omega_0 := \bigcup_n S_n$. Then $\Omega_0 \in \mathcal{S}$ (countable union of $\sigma$-finite sets is $\sigma$-finite) and $\|\psi|_{L^p(\Omega_0)}\| \ge \|\psi|_{L^p(S_n)}\| > M - 1/n$ for all $n$, so $\|\psi|_{L^p(\Omega_0)}\| = M$. Now suppose $f \in L^p(\mu)$. We show $\psi(f \mathbb{1}_{\Omega \setminus \Omega_0}) = 0$. The support $\{f \mathbb{1}_{\Omega \setminus \Omega_0} \ne 0\} \cap \Omega_0 = \varnothing$. Set $T_f := \{f \ne 0\} \setminus \Omega_0$. Since $f \in L^p$, the set $T_f$ is $\sigma$-finite (write $T_f = \bigcup_k \{1/(k+1) < |f| \le 1/k\} \cup \{|f| > 1\}$, each having finite measure: e.g. $\mu(\{|f| > 1\}) \le \int |f|^p \, d\mu < \infty$). Then $\Omega_0 \cup T_f \in \mathcal{S}$. By construction, $f \mathbb{1}_{\Omega \setminus \Omega_0} \in L^p(\Omega_0 \cup T_f)$ (it vanishes outside $T_f \subseteq \Omega_0 \cup T_f$) and also $f \mathbb{1}_{\Omega \setminus \Omega_0} \in L^p(T_f)$. We split: define $\psi_1, \psi_2 \in L^p(\Omega_0 \cup T_f)^*$ by $\psi_1(h) := \psi(h \mathbb{1}_{\Omega_0})$ and $\psi_2(h) := \psi(h \mathbb{1}_{T_f})$, so $\psi|_{L^p(\Omega_0 \cup T_f)} = \psi_1 + \psi_2$ on functions supported in $\Omega_0 \cup T_f$. By Step 11 (applied to $\sigma$-finite $\mu|_{\Omega_0 \cup T_f}$ and $p \in (1, \infty)$, $q \in (1, \infty)$), there is $g \in L^q(\mu|_{\Omega_0 \cup T_f})$ with $\psi(h) = \int_{\Omega_0 \cup T_f} h g \, d\mu$ for $h \in L^p(\Omega_0 \cup T_f)$, and the operator norm of this restriction equals $\|g\|_q$. Decompose $g = g_1 + g_2$ with $g_1 := g \mathbb{1}_{\Omega_0}$, $g_2 := g \mathbb{1}_{T_f}$. Then $\psi_j(h) = \int h g_j \, d\mu$ and disjointness of supports gives \begin{align*} \|\psi|_{L^p(\Omega_0 \cup T_f)}\|^q = \|g\|_q^q = \|g_1\|_q^q + \|g_2\|_q^q = \|\psi_1\|^q + \|\psi_2\|^q. \end{align*} Now $\|\psi_1\| = \|\psi|_{L^p(\Omega_0)}\| = M$ and $\|\psi|_{L^p(\Omega_0 \cup T_f)}\| \le M$ (by definition of $M$, since $\Omega_0 \cup T_f \in \mathcal{S}$). So $M^q \ge M^q + \|\psi_2\|^q$, forcing $\|\psi_2\| = 0$, i.e. $\psi_2 \equiv 0$, in particular $\psi(f \mathbb{1}_{\Omega \setminus \Omega_0}) = \psi_2(f) = 0$. Therefore $\psi(f) = \psi(f \mathbb{1}_{\Omega_0}) + \psi(f \mathbb{1}_{\Omega \setminus \Omega_0}) = \psi(f \mathbb{1}_{\Omega_0})$ for every $f \in L^p(\mu)$. Apply Step 11 on $\Omega_0$ to get $g_0 \in L^q(\mu|_{\Omega_0})$ with $\psi(f \mathbb{1}_{\Omega_0}) = \int_{\Omega_0} f g_0 \, d\mu$ for $f \in L^p(\Omega_0)$. Extend $g_0$ by zero on $\Omega \setminus \Omega_0$ to get $g \in L^q(\mu)$, then for any $f \in L^p(\mu)$: \begin{align*} \psi(f) = \psi(f \mathbb{1}_{\Omega_0}) = \int_{\Omega_0} f g_0 \, d\mu = \int_\Omega f g \, d\mu = \varphi_g(f). \end{align*} This proves surjectivity for general $\mu$ when $p \in (1, \infty)$. [guided] Why does the argument for general $\mu$ work only for $p \in (1, \infty)$ and not $p = 1$? The key step is the disjoint-support norm identity \begin{align*} \|g\|_q^q = \|g_1\|_q^q + \|g_2\|_q^q, \end{align*} which holds for $q \in [1, \infty)$, equivalently $p \in (1, \infty]$. The case $p = 1$ corresponds to $q = \infty$, and the identity becomes $\max(\|g_1\|_\infty, \|g_2\|_\infty) = \|g\|_\infty$ — which is true, but no longer additive in a way that lets us subtract $\|g_1\|_\infty^q = M^q$ from both sides to force $\|g_2\|_\infty = 0$. The whole "localise to $\sigma$-finite carrier" trick fails without additivity. This is consistent with the actual statement of the theorem: for $p = 1$, the duality $L^1(\mu)^* \cong L^\infty(\mu)$ requires $\mu$ to be $\sigma$-finite. (In fact, the duality genuinely fails for non-$\sigma$-finite measures: there are bounded functionals on $L^1$ for certain non-$\sigma$-finite measures that are not represented by any element of $L^\infty$.) For $p \in (1, \infty)$, the duality $L^p(\mu)^* \cong L^q(\mu)$ holds for any measure space because of the localisation argument above: the carrier $\Omega_0$ where $\psi$ "lives" is automatically $\sigma$-finite. [/guided] [/step] [step:Conclusion] Combining Steps 1-12: - $\varphi$ is linear (Step 1). - $\varphi$ is an isometry: $\|\varphi_g\| = \|g\|_q$ from the upper bound (Step 1) plus the lower bounds (Step 3 for $p \in (1, \infty)$, Step 4 for $p = 1$ with $\mu$ $\sigma$-finite). - $\varphi$ is surjective: Step 10 (finite $\mu$), Step 11 ($\sigma$-finite $\mu$), Step 12 (general $\mu$, $p > 1$). For (i), $p \in (1, \infty)$ and any measure $\mu$: $\varphi$ is an isometric isomorphism, so $L^p(\mu)^* \cong L^q(\mu)$. For (ii), $p = 1$ and $\sigma$-finite $\mu$: $\varphi$ is an isometric isomorphism, so $L^1(\mu)^* \cong L^\infty(\mu)$. This completes the proof. [/step]