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! \(\left\{{(x, t) {~\mathrel{\Big\vert}~}0\leq t \leq f(x)}\right\} = \left\{{ f(x) \geq t}\right\} \cap\left\{{ t\geq 0 }\right\}\)


  • 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.


  • 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