Write \begin{align*} \Gamma \coloneqq\left\{{(x, f(x)) {~\mathrel{\Big\vert}~}x\in {\mathbf{R}}}\right\} \subseteq {\mathbf{R}}^d .\end{align*} Then \begin{align*} \mu(\Gamma) &= \int_{{\mathbf{R}}^d} \chi_\Gamma \,d\mu\\ &= \int_{{\mathbf{R}}^{d-1}}\int_{\mathbf{R}}\chi_\Gamma(x, y) \,dy\,dx\\ &= \int_{{\mathbf{R}}^{d-1}} 0 \,dx\\ &= 0 ,\end{align*} using that \(\int_{\mathbf{R}}\chi_\Gamma(x, y) \,dy= 0\) since if \(x\) is fixed then \(\chi_\Gamma(x, y) = \left\{{f(x)}\right\}\) is a point with measure zero. Since \(f\) is measurable, \(\Gamma\) is a measurable set and \(\chi_\Gamma\) is measurable. Since the iterated integral was finite, the equalities are justified by Fubini-Tonelli.

• If these norms can be computed via iterated integrals, we have \begin{align*} {\left\lVert {f\ast g} \right\rVert}_1 &\coloneqq\int_{\mathbf{R}}{\left\lvert {(f\ast g)(x)} \right\rvert} \,dx\\ &\coloneqq\int_{\mathbf{R}}{\left\lvert {\int_{\mathbf{R}}H(x, y) \,dy} \right\rvert} \,dx\\ &\coloneqq\int_{\mathbf{R}}{\left\lvert {\int_{\mathbf{R}}f(y)g(x-y) \,dy} \right\rvert} \,dx\\ &\leq \int_{\mathbf{R}}\int_{\mathbf{R}}{\left\lvert {f(y) g(x-y)} \right\rvert} \,dx\,dy\\ &\coloneqq\int_{\mathbf{R}}\int_{\mathbf{R}}{\left\lvert {H(x ,y)} \right\rvert}\,dx\,dy\\ &\coloneqq\int_{{\mathbf{R}}^2} {\left\lvert {H} \right\rvert} \,d\mu_{{\mathbf{R}}^2} \\ &\coloneqq{\left\lVert {H} \right\rVert}_{L^1({\mathbf{R}}^2)} .\end{align*} So it suffices to show \({\left\lVert {H} \right\rVert}_1 < \infty\).

• A preliminary computation, the validity of which we will show afterward: \begin{align*} {\left\lVert {H} \right\rVert}_1 &\coloneqq{\left\lVert {H} \right\rVert}_{L^1({\mathbf{R}}^2)} \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)g(x-y)} \right\rvert} \, dy } \, dx && \text{Tonelli} \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)g(x-y)} \right\rvert} \, dx} \, dy && \text{Tonelli} \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)g(t)} \right\rvert} \, dt} \, dy && \text{setting } t=x-y, \,dt = - dx \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)} \right\rvert}\cdot {\left\lvert {g(t)} \right\rvert} \, dt}\, dy \\ &= \int _{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} \cdot \qty{ \int_{\mathbf{R}}{\left\lvert {g(t)} \right\rvert} \, dt}\, dy \\ &\coloneqq\int _{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} \cdot {\left\lVert {g} \right\rVert}_1 \,dy \\ &= {\left\lVert {g} \right\rVert}_1 \int _{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} \,dy &&\text{the norm is a constant} \\ &\coloneqq{\left\lVert {g} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 \\ &< \infty && \text{by assumption} .\end{align*}

• We’ve used Tonelli twice: to equate the integral to the iterated integral, and to switch the order of integration, so it remains to show the hypothesis of Tonelli are fulfilled.

• Apply Tonelli to \({\left\lvert {H} \right\rvert}\)
  • \(H\) measurable implies \({\left\lvert {H} \right\rvert}\) is measurable.
  • \({\left\lvert {H} \right\rvert}\) is non-negative.
  • So the iterated integrals are equal in the extended sense
  • The calculation shows the iterated integral is finite, so \(\int {\left\lvert {H} \right\rvert}\) is finite and \(H\) is thus integrable on \({\mathbf{R}}^2\).

Note: Fubini is not needed, since we’re not calculating the actual integral, just showing \(H\) is integrable.

Note: \(f\) is a function \({\mathbf{R}}\to {\mathbf{R}}\) in the original problem, but here I’ve assumed \(f:{\mathbf{R}}^n\to {\mathbf{R}}\).

• Since \(f\geq 0\), set \begin{align*} E\coloneqq\left\{{(x, t) \in {\mathbf{R}}^{n} \times{\mathbf{R}}{~\mathrel{\Big\vert}~}f(x) > t}\right\} = \left\{{(x, t) \in {\mathbf{R}}^n \times{\mathbf{R}}{~\mathrel{\Big\vert}~}0 \leq t < f(x)}\right\} .\end{align*}

• Claim: since \(f\) is measurable, \(E\) is measurable and thus \(m(E)\) makes sense.

  • Since \(f\) is measurable, \(F(x, t) \coloneqq t - f(x)\) is measurable on \({\mathbf{R}}^n \times{\mathbf{R}}\).
  • Then write \(E = \left\{{F < 0}\right\} \cap\left\{{t\geq 0}\right\}\) as an intersection of measurable sets.

• We have slices \begin{align*} E^t &\coloneqq\left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}(x, t) \in E}\right\} = \left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}0 \leq t < f(x)}\right\} \\ E^x &\coloneqq\left\{{t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}(x, t) \in E}\right\} = \left\{{t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}0 \leq t \leq f(x)}\right\} = [0, f(x)] .\end{align*}

  • \(E_t\) is precisely the set that appears in the original RHS integrand.
  • \(m(E^x) = f(x)\).

• Claim: \(\chi_E\) satisfies the conditions of Tonelli, and thus \(m(E) = \int \chi_E\) is equal to any iterated integral.

  • Non-negative: clear since \(0\leq \chi_E \leq 1\)
  • Measurable: characteristic functions of measurable sets are measurable.

• Conclude:

  • For almost every \(x\), \(E^x\) is a measurable set, \(x\mapsto m(E^x)\) is a measurable function, and \(m(E) = \int_{{\mathbf{R}}^n} m(E^x) \, dx\)
  • For almost every \(t\), \(E^t\) is a measurable set, \(t\mapsto m(E^t)\) is a measurable function, and \(m(E) = \int_{{\mathbf{R}}} m(E^t) \, dt\)

• On one hand, \begin{align*} m(E) &= \int_{{\mathbf{R}}^{n+1}} \chi_E(x, t) \\ &= \int_{{\mathbf{R}}} \int_{{\mathbf{R}}^n} \chi_E(x, t) \,dt \,dx \quad\text{by Tonelli}\\ &= \int_{{\mathbf{R}}^n} m(E^x) \,dx \quad\text{first conclusion}\\ &= \int_{{\mathbf{R}}^n} f(x) \,dx .\end{align*}

• On the other hand, \begin{align*} m(E) &= \int_{{\mathbf{R}}^{n+1}} \chi_E(x, t) \\ &= \int_{\mathbf{R}}\int_{{\mathbf{R}}^n} \chi_E(x, t) \, dx \,dt \quad\text{by Tonelli} \\ &= \int_{\mathbf{R}}m(E^t) \,dt \quad\text{second conclusion} .\end{align*}

• Thus \begin{align*} \int_{{\mathbf{R}}^n} f \,dx = m(E) = \int_{\mathbf{R}}m(E^t) \,dt = \int_{\mathbf{R}}m\qty{\left\{{x{~\mathrel{\Big\vert}~}f(x) > t}\right\}} .\end{align*}