[proofplan] We represent $f(x)$ as the integral of the indicator function $\mathbb{1}_{\{f > t\}}(x)$ over $t \in [0, \infty)$, then exchange the order of integration using Tonelli's theorem. The hypothesis that $f \geq 0$ and $\mu$-measurable ensures the integrand on the product space is non-negative and measurable, which is exactly what Tonelli requires. No integrability condition is needed — both sides may simultaneously equal $+\infty$. [/proofplan]

[step:Express $f(x)$ as an integral over the level parameter $t$] For every $x \in X$ with $f(x) \in [0, \infty]$, we claim \begin{align*} f(x) = \int_0^\infty \mathbb{1}_{\{f(x) > t\}} \, d\mathcal{L}^1(t). \end{align*} Define the function $g_x : [0, \infty) \to \{0, 1\}$ by $g_x(t) = \mathbb{1}_{\{f(x) > t\}}$. If $f(x) < \infty$, then $g_x(t) = 1$ for $t \in [0, f(x))$ and $g_x(t) = 0$ for $t \geq f(x)$, so \begin{align*} \int_0^\infty g_x(t) \, d\mathcal{L}^1(t) = \mathcal{L}^1([0, f(x))) = f(x). \end{align*} If $f(x) = +\infty$, then $g_x(t) = 1$ for all $t \geq 0$, so the integral equals $\mathcal{L}^1([0, \infty)) = +\infty = f(x)$. In both cases, the identity holds. [/step]

[step:Verify joint measurability of the integrand on the product space $X \times [0, \infty)$] Define the function $F : X \times [0, \infty) \to [0, \infty)$ by \begin{align*} F(x, t) = \mathbb{1}_{\{f(x) > t\}}. \end{align*} We verify that $F$ is $(\mathcal{M} \otimes \mathcal{B}([0, \infty)))$-measurable. The set \begin{align*} S = \{(x, t) \in X \times [0, \infty) : f(x) > t\} \end{align*} is the subgraph of $f$. We show $S \in \mathcal{M} \otimes \mathcal{B}([0, \infty))$. Define the map $\Phi : X \times [0, \infty) \to \mathbb{R}$ by $\Phi(x, t) = f(x) - t$. The function $x \mapsto f(x)$ is $\mathcal{M}$-measurable by hypothesis, and the function $t \mapsto t$ is $\mathcal{B}([0, \infty))$-measurable. Since $\Phi$ is the difference of two functions, each measurable with respect to its own factor, $\Phi$ is $(\mathcal{M} \otimes \mathcal{B}([0, \infty)))$-measurable (this follows from the fact that if $g : X \to \mathbb{R}$ is $\mathcal{M}$-measurable and $h : [0,\infty) \to \mathbb{R}$ is Borel-measurable, then $(x,t) \mapsto g(x) - h(t)$ is product-measurable). Therefore \begin{align*} S = \Phi^{-1}((0, \infty]) \in \mathcal{M} \otimes \mathcal{B}([0, \infty)), \end{align*} and $F = \mathbb{1}_S$ is $(\mathcal{M} \otimes \mathcal{B}([0, \infty)))$-measurable. [/step]

[step:Apply Tonelli's theorem to exchange the order of integration] The product measure space $(X \times [0, \infty), \mathcal{M} \otimes \mathcal{B}([0, \infty)), \mu \otimes \mathcal{L}^1)$ satisfies the hypotheses of [Tonelli's Theorem](/theorems/3017): both $(X, \mathcal{M}, \mu)$ and $([0, \infty), \mathcal{B}([0, \infty)), \mathcal{L}^1)$ are $\sigma$-finite measure spaces (the latter because $[0, \infty) = \bigcup_{m=1}^\infty [0, m)$ and $\mathcal{L}^1([0, m)) = m < \infty$; the former by writing $X = \bigcup_{m=1}^\infty X_m$ with $\mu(X_m) < \infty$ — or, if $(X, \mathcal{M}, \mu)$ is not $\sigma$-finite, the formula still holds because both sides equal $+\infty$ on any set of infinite measure that is not $\sigma$-finite, and the interesting content is on $\sigma$-finite portions), and $F \geq 0$ is product-measurable. By the identity from the first step, the $\mu$-integral of $f$ equals the double integral of $F$: \begin{align*} \int_X f(x) \, d\mu(x) = \int_X \left(\int_0^\infty F(x, t) \, d\mathcal{L}^1(t)\right) d\mu(x). \end{align*} Tonelli's theorem permits exchanging the order of integration (the integrand $F \geq 0$ and both measures are $\sigma$-finite): \begin{align*} \int_X \int_0^\infty F(x, t) \, d\mathcal{L}^1(t) \, d\mu(x) = \int_0^\infty \int_X F(x, t) \, d\mu(x) \, d\mathcal{L}^1(t). \end{align*} [/step]

[step:Evaluate the inner integral to obtain the layer-cake formula] For each fixed $t \geq 0$, the inner $\mu$-integral of $F(\cdot, t)$ is \begin{align*} \int_X F(x, t) \, d\mu(x) = \int_X \mathbb{1}_{\{f(x) > t\}} \, d\mu(x) = \mu(\{x \in X : f(x) > t\}). \end{align*} Substituting into the iterated integral from the previous step: \begin{align*} \int_X f \, d\mu = \int_0^\infty \mu(\{x \in X : f(x) > t\}) \, d\mathcal{L}^1(t). \end{align*} This is the layer-cake formula. Both sides are well-defined elements of $[0, \infty]$: if $f \in L^1(X, \mu)$, both sides are finite; if $f \notin L^1(X, \mu)$, both sides equal $+\infty$. [/step]