# Fubini-Tonelli

## Spring 2021.6 #real_analysis/qual/work

This problem may be much harder than expected. Recommended skip.

Let $$f: {\mathbf{R}}\times{\mathbf{R}}\to {\mathbf{R}}$$ be a measurable function and for $$x\in {\mathbf{R}}$$ define the set \begin{align*} E_x \coloneqq\left\{{ y\in {\mathbf{R}}{~\mathrel{\Big\vert}~}\mu\qty{ z\in {\mathbf{R}}{~\mathrel{\Big\vert}~}f(x,z) = f(x, y) } > 0 }\right\} .\end{align*} Show that the following set is a measurable subset of $${\mathbf{R}}\times{\mathbf{R}}$$: \begin{align*} E \coloneqq\bigcup_{x\in {\mathbf{R}}} \left\{{ x }\right\} \times E_x .\end{align*}

Hint: consider the measurable function $$h(x,y,z) \coloneqq f(x, y) - f(x, z)$$.

## Fall 2021.4 #real_analysis/qual/completed

Let $$f$$ be a measurable function on $$\mathbb{R}$$. Show that the graph of $$f$$ has measure zero in $$\mathbb{R}^{2}$$.

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.

## Spring 2020.4 #real_analysis/qual/completed

Let $$f, g\in L^1({\mathbf{R}})$$. Argue that $$H(x, y) \coloneqq f(y) g(x-y)$$ defines a function in $$L^1({\mathbf{R}}^2)$$ and deduce from this fact that \begin{align*} (f\ast g)(x) \coloneqq\int_{\mathbf{R}}f(y) g(x-y) \,dy \end{align*} defines a function in $$L^1({\mathbf{R}})$$ that satisfies \begin{align*} {\left\lVert {f\ast g} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1 .\end{align*}

Just do it! Sort out the justification afterward. Use Tonelli.

• Tonelli: non-negative and measurable yields measurability of slices and equality of iterated integrals
• Fubini: $$f(x, y) \in L^1$$ yields integrable slices and equality of iterated integrals
• F/T: apply Tonelli to $${\left\lvert {f} \right\rvert}$$; if finite, $$f\in L^1$$ and apply Fubini to $$f$$
• See Folland’s Real Analysis II, p. 68 for a discussion of using Fubini and 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.

$$H$$ is measurable on $${\mathbf{R}}^2$$:

• It suffices to show $$\tilde f(x, y) \coloneqq f(y)$$ and $$\tilde g(x, y) \coloneqq g(x-y)$$ are both measurable on $${\mathbf{R}}^2$$.
• Then use that products of measurable functions are measurable.
• $$f \in L^1$$ by assumption, and $$L^1$$ functions are measurable by definition.
• The function $$(x, y) \mapsto g(x-y)$$ is measurable on $${\mathbf{R}}^2$$:
• $$g$$ is measurable on $${\mathbf{R}}$$ by assumption, so the cylinder function $$G(x, y) \coloneqq g(x)$$ on $${\mathbf{R}}^2$$ is measurable (result from course).
• # Define a linear transformation  \begin{align*} T \coloneqq \begin{bmatrix} 1 & -1 \ 0 & 1 \end{bmatrix} \in \operatorname{GL}_2({\mathbf{R}}) && \implies ,,, T \begin{bmatrix} x \ y \end{bmatrix}

\begin{bmatrix} x-y
\ y
\end{bmatrix} ,\end{align*} {=html} and linear functions are measurable.
• Write \begin{align*} \tilde g(x-y) \coloneqq G(x-y, y) \coloneqq(G\circ T)(x, y) ,\end{align*} and compositions of measurable functions are measurable.
• 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.

## Spring 2019.4 #real_analysis/qual/completed

Let $$f$$ be a non-negative function on $${\mathbf{R}}^n$$ and $$\mathcal A = \{(x, t) ∈ {\mathbf{R}}^n \times {\mathbf{R}}: 0 ≤ t ≤ f (x)\}$$.

Prove the validity of the following two statements:

• $$f$$ is a Lebesgue measurable function on $${\mathbf{R}}^n \iff \mathcal A$$ is a Lebesgue measurable subset of $${\mathbf{R}}^{n+1}$$

• If $$f$$ is a Lebesgue measurable function on $${\mathbf{R}}^n$$, then \begin{align*} m(\mathcal{A})=\int _{{\mathbf{R}}^{n}} f(x) d x=\int_{0}^{\infty} m\left(\left\{x \in {\mathbf{R}}^{n}: f(x) \geq t\right\}\right) dt \end{align*}

• See Stein and Shakarchi p.82 corollary 3.3.
• Tonelli
• Important trick! $$\left\{{(x, t) {~\mathrel{\Big\vert}~}0\leq t \leq f(x)}\right\} = \left\{{ f(x) \geq t}\right\} \cap\left\{{ t\geq 0 }\right\}$$

$$\implies$$:

• Suppose $$f:{\mathbf{R}}^n\to {\mathbf{R}}$$ is a measurable function.
• Rewrite $$A$$: \begin{align*} A &= \left\{{ (x, t) \in {\mathbf{R}}^d \times{\mathbf{R}}{~\mathrel{\Big\vert}~}0\leq t \leq f(x) }\right\} \\ &= \left\{{ (x, t) \in {\mathbf{R}}^d \times{\mathbf{R}}{~\mathrel{\Big\vert}~}0 \leq t < \infty }\right\} \cap\left\{{ (x, t) \in {\mathbf{R}}^d\times{\mathbf{R}}{~\mathrel{\Big\vert}~}t\leq f(x) }\right\} \\ &= \qty{ {\mathbf{R}}^d \times[0, \infty) } \cap\left\{{ (x, t) \in {\mathbf{R}}^d\times{\mathbf{R}}{~\mathrel{\Big\vert}~}f(x) -t \geq 0 }\right\} \\ &\coloneqq\qty{ {\mathbf{R}}^d \times[0, \infty) } \cap H^{-1}\qty{[0, \infty)} ,\end{align*} where we define \begin{align*} H: {\mathbf{R}}^d \times{\mathbf{R}}&\to {\mathbf{R}}\\ (x, t) &\mapsto f(x) - t .\end{align*}
• Note: this is “clearly” measurable!
• If we can show both sets are measurable, we’re done, since $$\sigma{\hbox{-}}$$algebras are closed under countable intersections.
• The first set is measurable since it is a Borel set in $${\mathbf{R}}^{d+1}$$.
• For the same reason, it suffices to show $$H$$ is a measurable function.
• Define cylinder functions \begin{align*} F: {\mathbf{R}}^d \times{\mathbf{R}}&\to {\mathbf{R}}\\ (x, t) &\mapsto f(x) \end{align*} and \begin{align*} G: {\mathbf{R}}^d \times{\mathbf{R}}&\to {\mathbf{R}}\\ (x, t) &\mapsto t \end{align*}
• $$F$$ is a cylinder of $$f$$, and since $$f$$ is measurable by assumption, $$F$$ is measurable.
• $$G$$ is a cylinder on the identity for $${\mathbf{R}}$$, which is measurable, so $$G$$ is measurable.
• Define \begin{align*} H: {\mathbf{R}}^d &\to {\mathbf{R}}\\ (x, t) &\mapsto F(x, t) - G(x, t) \coloneqq f(x) - t ,\end{align*} which are linear combinations of measurable functions and thus measurable.

$$\impliedby$$:

• Suppose $${\mathcal{A}}$$ is a measurable set.

• A corollary of Tonelli applied to $$\chi_X$$: if $$E$$ is measurable, then for a.e. $$t$$ the following slice is measurable: \begin{align*} {\mathcal{A}}_t \coloneqq\left\{{ x \in {\mathbf{R}}^d {~\mathrel{\Big\vert}~}(x,t) \in {\mathcal{A}}}\right\} &= \left\{{x\in {\mathbf{R}}^d {~\mathrel{\Big\vert}~}f(x) \geq t \geq 0}\right\} \\ &= f^{-1}\qty{[t, \infty)} .\end{align*}

• But maybe this isn’t enough, because we need $$f^{-1}\qty{[\alpha, \infty)}$$ for all $$\alpha$$
• But the other slice is also measurable for a.e. $$x$$: \begin{align*} {\mathcal{A}}_x &\coloneqq\left\{{ t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}(x, t) \in {\mathcal{A}}}\right\} \\ &= \left\{{ t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}0 \leq t \leq f(x) }\right\} \\ &= \left\{{ t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}t\in [0, f(x)] }\right\} \\ &= [0, f(x)] .\end{align*}

• Moreover the function $$x\mapsto m({\mathcal{A}}_x)$$ is a measurable function of $$x$$

• Now note $$m({\mathcal{A}}_x) = f(x) - 0 = f(x)$$, so $$f$$ must be measurable.

• Writing down what the slices are \begin{align*} \mathcal{A} &= \left\{{(x, t) \in {\mathbf{R}}^n\times{\mathbf{R}}{~\mathrel{\Big\vert}~}0 \leq t \leq f(x)}\right\} \\ \mathcal{A}_t &= \left\{{x \in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}t\leq f(x) }\right\} .\end{align*}

• Then \begin{align*} \int_{{\mathbf{R}}^n} f(x) ~dx &= \int_{{\mathbf{R}}^n} \int_0^{f(x)} 1 ~dt~dx \\ &= \int_{{\mathbf{R}}^n} \int_{0}^\infty \chi_\mathcal{A} ~dt~dx \\ &\overset{F.T.}= \int_{0}^\infty \int_{{\mathbf{R}}^n} \chi_\mathcal{A} ~dx~dt\\ &= \int_0^\infty m(\mathcal{A}_t) ~dt ,\end{align*} where we just use that $$\int \int \chi_\mathcal{A} = m(\mathcal{A})$$

• By Tonelli, all of these integrals are equal.

• This is justified because $$f$$ was assumed measurable on $${\mathbf{R}}^n$$, thus by (a) $$\mathcal{A}$$ is a measurable set and thus $$\chi_A$$ is a measurable function on $${\mathbf{R}}^n\times{\mathbf{R}}$$.

## Fall 2018.5 #real_analysis/qual/completed

Let $$f \geq 0$$ be a measurable function on $${\mathbf{R}}$$. Show that \begin{align*} \int _{{\mathbf{R}}} f = \int _{0}^{\infty} m(\{x: f(x)>t\}) dt \end{align*}

• Claim: If $$E\subseteq {\mathbf{R}}^a \times{\mathbf{R}}^b$$ is a measurable set, then for almost every $$y\in {\mathbf{R}}^b$$, the slice $$E^y$$ is measurable and \begin{align*} m(E) = \int_{{\mathbf{R}}^b} m(E^y) \,dy .\end{align*}
• Set $$g = \chi_E$$, which is non-negative and measurable, so apply Tonelli.
• Conclude that $$g^y = \chi_{E^y}$$ is measurable, the function $$y\mapsto \int g^y(x)\, dx$$ is measurable, and $$\int \int g^y(x)\,dx \,dy = \int g$$.
• But $$\int g = m(E)$$ and $$\int\int g^y(x) \,dx\,dy = \int m(E^y)\,dy$$.

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*}

## Fall 2015.5 #real_analysis/qual/completed

Let $$f, g \in L^1({\mathbf{R}})$$ be Borel measurable.

• Show that
• The function \begin{align*}F(x, y) \coloneqq f(x-y) g(y)\end{align*} is Borel measurable on $${\mathbf{R}}^2$$, and
• For almost every $$x\in {\mathbf{R}}$$, the function $$f(x-y)g(y)$$ is integrable with respect to $$y$$ on $${\mathbf{R}}$$.
• Show that $$f\ast g \in L^1({\mathbf{R}})$$ and \begin{align*} \|f * g\|_{1} \leq \|f\|_{1} \|g\|_{1} \end{align*}

• $$F \in {\mathcal{B}}({\mathbf{R}}^2)$$:

• Write a function $$\tilde f(x, y) \coloneqq f(x)$$
• Write a linear transformation $$T = { \begin{bmatrix} {1} & {0} \\ {0} & {-1} \end{bmatrix} } \in \operatorname{GL}_2$$, so $$T{\left[ {x, y} \right]} = {\left[ {x-y, 0} \right]}$$
• Write $$f(x-y) \coloneqq(\tilde f \circ T)(x, y)$$, which is a composition of measurable functions and thus measurable.
• A product of measurable functions is measurable.
• $$f\ast g \in L^1({\mathbf{R}})$$: estimate \begin{align*} \int {\left\lvert { f\ast g} \right\rvert} d\mu &= \int_{\mathbf{R}}\int_{\mathbf{R}}{\left\lvert {f(x-y)g(y)} \right\rvert}\,dx\,dy\\ &= \int_{\mathbf{R}}\int_{\mathbf{R}}{\left\lvert {f(x-y)} \right\rvert}{\left\lvert {g(y)} \right\rvert}\,dx\,dy\\ &= \int_{\mathbf{R}}{\left\lvert {g(y)} \right\rvert} \int_{\mathbf{R}}{\left\lvert {f(x-y)} \right\rvert}\,dx\,dy\\ &= {\left\lVert {g} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 ,\end{align*} where we’ve used translation invariance of the $$L^1$$ norm and Fubini-Tonelli justified by the finite result.

• $$F_x(y) \coloneqq f(x-y)g(y)$$ is integrable with respect to $$y$$ for almost every $$x$$:

• This follows from Fubini-Tonelli, which says that if $$F(x, y)$$ is integrable, the slices $$F^x(y)$$ are integrable for almost every $$x$$. Here take $$F(x, y) \coloneqq f(x-y)g(y)$$.

## Spring 2014.5 #real_analysis/qual/work

Let $$f, g \in L^1([0, 1])$$ and for all $$x\in [0, 1]$$ define \begin{align*} F(x) \coloneqq\int _{0}^{x} f(y) \, dy {\quad \operatorname{and} \quad} G(x)\coloneqq\int _{0}^{x} g(y) \, dy. \end{align*}

Prove that \begin{align*} \int _{0}^{1} F(x) g(x) \, dx = F(1) G(1) - \int _{0}^{1} f(x) G(x) \, dx \end{align*}

#real_analysis/qual/work #real_analysis/qual/completed