Fubini-Tonelli

Spring 2021.6 #real_analysis/qual/work

warnings:

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

problem (?):

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

solution:

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 $\begin{align*} {\left\lVert {f\ast g} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1 .\end{align*}$

strategy:

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

concept:

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.

solution:

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.

claim:

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

proof (?):

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 $\begin{align*} m(\mathcal{A})=\int _{{\mathbf{R}}^{n}} f(x) d x=\int_{0}^{\infty} m\left(\left\{x \in {\mathbf{R}}^{n}: f(x) \geq t\right\}\right) dt \end{align*}$

concept:

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

solution:

proof (a, $\implies$):

$\implies$ :

Suppose $f:{\mathbf{R}}^n\to {\mathbf{R}}$ is a measurable function.
Rewrite $A$ :
A={(x,t)∈Rd×R | 0≤t≤f(x)}={(x,t)∈Rd×R | 0≤t<∞}∩{(x,t)∈Rd×R | t≤f(x)}=(Rd×[0,∞))∩{(x,t)∈Rd×R | f(x)−t≥0}:=(Rd×[0,∞))∩H−1([0,∞)), 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
F:Rd×R→R(x,t)↦f(x) 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.

proof (a, $\impliedby$):

$\impliedby$ :

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.

proof (of b):

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 $\begin{align*} \int _{{\mathbf{R}}} f = \int _{0}^{\infty} m(\{x: f(x)>t\}) dt \end{align*}$

concept:

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
m(E)=∫Rbm(Ey)dy.
- 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 $\int\int g^y(x) \,dx\,dy = \int m(E^y)\,dy$ .

solution:

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

problem (?):

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 $\begin{align*} \|f * g\|_{1} \leq \|f\|_{1} \|g\|_{1} \end{align*}$

solution:

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

problem (?):

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

Fubini-Tonelli

Spring 2021.6 #real_analysis/qual/work

Fall 2021.4 #real_analysis/qual/completed

Spring 2020.4 #real_analysis/qual/completed

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}

Spring 2019.4 #real_analysis/qual/completed

Fall 2018.5 #real_analysis/qual/completed

Fall 2015.5 #real_analysis/qual/completed

Spring 2014.5 #real_analysis/qual/work