**Step 1: The map $\Phi$ is a well-defined isometry.** For $g \in L^q$, the functional $\Phi_g(f) := \int fg \, d\mu$ satisfies $|\Phi_g(f)| \le \|f\|_p \|g\|_q$ by the [Hölder Inequality](/theorems/516), so $\|\Phi_g\| \le \|g\|_q$. Taking $f = \operatorname{sign}(g) |g|^{q-1} / \|g\|_q^{q/p}$ (which lies in $L^p$ since $|f|^p = |g|^{(q-1)p} / \|g\|_q^q = |g|^q / \|g\|_q^q$, and $q = p(q-1)$) gives $\Phi_g(f) = \|g\|_q$, so $\|\Phi_g\| = \|g\|_q$. **Step 2: Surjectivity for $p \in (1, \infty)$.** The image $M := \Phi(L^q)$ is a closed subspace of $(L^p)^*$ (closed because $\Phi$ is an isometry: if $\Phi(g_n) \to F$ in $(L^p)^*$, then $(g_n)$ is Cauchy in $L^q$, so $g_n \to g \in L^q$ and $\Phi(g) = F$). Suppose $M \subsetneq (L^p)^*$. Since $L^p$ is reflexive (by the [Milman-Pettis Theorem](/theorems/899) and [Clarkson's Inequalities](/theorems/900)), $(L^p)^*$ is reflexive, so $M$ is reflexive as a closed subspace. If $\Psi \in (L^p)^{**}$ vanishes on $M$, reflexivity gives $\Psi = \hat{h}$ for some $h \in L^p$. Then $\hat{h}(\Phi_g) = \Phi_g(h) = \int hg \, d\mu = 0$ for all $g \in L^q$. Taking $g = |h|^{p-2}h \in L^q$ gives $\int |h|^p \, d\mu = 0$, so $h = 0$ and $\Psi = 0$. By Hahn-Banach, $M$ is dense, hence $M = (L^p)^*$. **Step 3: The case $p = 1$.** For $F \in (L^1)^*$, define $G: L^2 \to \mathbb{R}$ by $G(h) := F(\omega h)$ where $\omega(x) = (1 + |x|)^{-1}$. Then $|G(h)| \le \|F\| \cdot \|\omega\|_{L^2} \cdot \|h\|_{L^2}$ by Hölder, so $G \in (L^2)^*$. By the [Riesz Representation Theorem](/theorems/221) for [Hilbert spaces](/page/Hilbert%20Space), $G(h) = \int vh \, d\mu$ for some $v \in L^2$. Setting $g := v/\omega$ and using the representation on the dense subspace $\omega L^2 \supseteq C_c^\infty$, we get $F(f) = \int fg \, d\mu$ for all $f \in L^1$. A Chebyshev-type argument shows $g \in L^\infty$ with $\|g\|_\infty \le \|F\|$.