Math Problems Vault

Problem 6.4.1

(a)(b)

(L10) If $f \in L^{1}(E)$ and $\left|g\right| \leqslant f \mu$ -a. e. on $E$ , then $g \in L^{1}(E)$ . Also, any $f$ that is bounded $\mu$ -a. e. on a set $E$ with $\mu (E)<\infty$ and is zero $\mu -$ a. e. on $E^{c}$ is in $L^{1}(E)$ . (L16) We have the following, where $A^{\prime } \in \mathcal{F}^{\prime }$ :

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(a)

If $\varphi \geqslant 0$ , then $\int_{f^{-1}\left(A^{\prime }\right)} \varphi \circ f \mathrm{~ d} \mu =\int_{A^{\prime }} \varphi \mathrm{d}\left(\mu \circ f^{-1}\right)$ .

(b)

For general $\varphi , \int_{f^{-1}\left(A^{\prime }\right)} \varphi \circ f \mathrm{~ d} \mu$ exists and is finite iff $\int_{A^{\prime }} \varphi \mathrm{d}\left(\mu \circ f^{-1}\right)$ exists and is finite, and in this case equality obtains.

Problem 6.4.2

(a)(b)(c)

Suppose that $\sum_{n=1}^{\infty } f_{n}$ converges $\mu$ -a.e. on $\Omega$ and that for every $n \in \mathbb {N}$ we have $\left|\sum_{k=1}^{n} f_{k}\right| \leqslant g$ $\mu$ -a.e. on $\Omega$ , where $g \in L^{1}(\Omega )$ .

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(a)

For every $n \in \mathbb {N}$ we have $f_{n} \in L^{1}(\Omega )$ , and $\sum_{n=1}^{\infty } f_{n} \in L^{1}(\Omega )$ as well.

(b)

The equality $\int \sum_{n=1}^{\infty } f_{n}=\sum_{n=1}^{\infty } \int f_{n}$ obtains.

(c)

Prove the following corollary: if $\sum_{n=1}^{\infty } \int \left|f_{n}\right|$ converges, then $\sum_{n=1}^{\infty } f_{n}$ converges absolutely $\mu$ -a.e. on $\Omega$ and again the equality $\int \sum_{n=1}^{\infty } f_{n}=\sum_{n=1}^{\infty } \int f_{n}$ holds.

Problem 6.4.3

(a)(b)

Suppose that $f \in L^{1}(\Omega )$ .

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(a)

Let $A_{n}=\left\{ \omega \in \Omega : \left|f(\omega )\right| \geqslant n\right\}$ . Then $\lim_{n}\left|f_{n}(\omega )\right| \; \mathbb {1}_{A_{n}}(\omega )=0$ for $\mu$ -a.e. $\omega \in \Omega$ .

(b)

Use DCT to show that $\lim_{n} \int_{A_{n}}\left|f\right|=0$ .

Problem 6.4.4

This exercise deals with the concept of uniform integrability. A sequence $\left\{ f_{n}\right\}_{n=1}^{\infty }$ is called uniformly integrable iff $\lim_{m} \sup_{n} \int_{A_{m}}\left|f_{n}\right|=0$ , where the sets $A_{1}, A_{2}, \ldots$ are as given in the previous exercise: $A_{n}=\left\{ \omega \in \Omega : \left|f(\omega )\right| \geqslant n\right\}$ for $f \in L^{1}(\Omega )$ .

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(a)

Suppose that $\mu (\Omega )<+\infty$ and $\left\{ f_{n}\right\}_{n=1}^{\infty }$ is uniformly integrable. Then each $f_{n}$ is in $L^{1}(\Omega )$ .

(b)

If $\left\{ f_{n}\right\}_{n=1}^{\infty }$ and $\left\{ g_{n}\right\}_{n=1}^{\infty }$ are uniformly integrable, then so is $\left\{ f_{n}+g_{n}\right\}_{n=1}^{\infty }$ .

(c)

Suppose that $\mu (\Omega )<+\infty$ , $f_{n} \rightarrow f$ $\mu$ -a.e. on $\Omega$ , and $\left\{ f_{n}\right\}_{n=1}^{\infty }$ is uniformly integrable. Then $f \in L^{1}(\Omega )$ and $\lim \int f_{n}=\int f$ .

(d)

Suppose that $\mu (\Omega )<+\infty$ and that $f_{n} \rightarrow f$ $\mu$ -a.e. on $\Omega$ , where each of $f, f_{1}, f_{2}, \ldots$ is nonnegative and in $L^{1}(\Omega )$ . If $\lim \int f_{n}=\int f$ , then $\left\{ f_{n}\right\}_{n=1}^{\infty }$ is uniformly integrable.

(e)

Suppose that $\mu (\Omega )<+\infty$ , $f, f_{1}, f_{2}, \ldots \in L^{1}(\Omega )$ , and $f_{n} \rightarrow f$ $\mu$ -a.e. on $\Omega$ . Then the following statements are logically equivalent:

$\left\{ f_{n}\right\}_{n=1}^{\infty }$ is uniformly integrable.
$\lim \int \left|f_{n}-f\right|=0$ .
$\lim \int \left|f_{n}\right| \rightarrow \int \left|f\right|$ .

(f)

Suppose that $\mu (\Omega )<+\infty$ . Then $\left\{ f_{n}\right\}_{n=1}^{\infty }$ is uniformly integrable iff (i) $\int \left|f_{n}\right|$ is bounded and (ii) for every $\varepsilon >0$ there is $\delta >0$ such that $\mu (E)<\delta$ implies $\int_{E}\left|f_{n}\right|<\varepsilon$ for every $n \in \mathbb {N}$ .

Remark.

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The reader may consult Billingsley's Probability and Measure, Section 16, for the solutions to (a)-(e).

Problem 6.4.5

(a)(b)

This problem deals with parametric integrals. Here, we consider a measure space $(\Omega , \mathcal{F}, \mu )$ as well as a nondegenerate interval $\Theta \subseteq \mathbb {R}$ of any form. Suppose that for every $\theta \in \Theta$ there is a function $f_{\theta }$ in $L^{1}(\Omega , \mathcal{F}, \mu )$ with domain $\Omega$ . That is, for every $\theta \in \Theta$ we may consider the real number $\int_{\Omega } f_{\theta }(\omega ) \, d\mu (\omega )$ . For each $\theta \in \Theta$ , write $F(\theta )$ for $\int_{\Omega } f_{\theta }(\omega ) \, d\mu (\omega )$ . Hold this notation and these assumptions fixed throughout.

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(a)

Make the following additional assumptions:

There exists $g \in L^{1}(\Omega , \mathcal{F}, \mu )$ with domain $\Omega$ such that $\left|f_{\theta }(\omega )\right| \leqslant g(\omega )$ for every $\omega \in \Omega$ and $\theta \in \Theta$ .
The element $\theta_{0} \in \Theta$ is such that $\lim_{\theta \rightarrow \theta_{0}} f_{\theta }(\omega )=f_{\theta_{0}}(\omega )$ for $\mu$ -a.e. $\omega \in \Omega$ . Show that $\lim_{\theta \rightarrow \theta_{0}} F(\theta )=F\left(\theta_{0}\right)$ , so that $F$ is continuous at $\theta_{0}$ .

(b)

Forget about the assumptions from (a), and instead make the following new set of assumptions:

For $\mu$ -a.e. $\omega \in \Omega$ we have that $f_{\theta }(\omega )$ is differentiable at each $\theta \in \Theta$ .
There exists $g \in L^{1}(\Omega , \mathcal{F}, \mu )$ with domain $\Omega$ such that $\left|\frac{\partial }{\partial \theta } f_{\theta }(\omega )\right| \leqslant g(\omega )$ for $\mu$ -a.e. $\omega \in \Omega$ and all $\theta \in \Theta$ . Show that (i) $F$ is differentiable on $\Theta$ , and (ii) $\frac{\partial }{\partial \theta } F(\theta )=\int_{\Omega } \frac{\partial }{\partial \theta } f_{\theta }(\omega ) \, d\mu (\omega )$ .

Problem 6.4.6

Assume that $f$ is an $\mathcal{F}/\mathcal{B}^{*}$ -measurable function defined $\mu$ -a.e. on $\Omega$ such that there are real numbers $m$ and $M$ with $m \leqslant f \leqslant M$ $\mu$ -a.e. on $\Omega$ . Also, assume that $g \in L^{1}(\Omega , \mathcal{F}, \mu )$ . Then there is $\alpha \in [m, M]$ such that $\int f\left|g\right|=\alpha \int \left|g\right|$ .

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Problem 6.4.7

Let $\left\{ f_{n}\right\}_{n=1}^{\infty }$ denote a sequence in $\mathfrak {N}$ with $f_{1} \geqslant f_{2} \geqslant \cdots \geqslant 0$ , $f_{n} \rightarrow f$ on $\Omega$ , and $f_{1} \in L^{1}(\Omega , \mathcal{F}, \mu )$ . Then $\lim_{n} \int f_{n}=\int f$ , and the requirement that $f_{1} \in L^{1}(\Omega , \mathcal{F}, \mu )$ is essential. [This mimics MCT somewhat.]

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Problem 6.4.8

Suppose that $\mu (\Omega )<+\infty$ , and let $\left\{ f_{n}\right\}_{n=1}^{\infty }$ denote a uniformly bounded sequence of functions on $\Omega$ with $f_{n} \rightarrow f$ uniformly on $\Omega$ . Then $\lim_{n} \int f_{n}=\int f$ .

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Problem 6.4.9

Suppose that $\left\{ f_{n}\right\}_{n=1}^{\infty }$ is a sequence of functions in $L^{1}(\Omega , \mathcal{F}, \mu )$ such that the series $\sum_{n=1}^{\infty } \int \left|f_{n}\right|$ converges. Then $\sum_{n=1}^{\infty } f_{n}$ converges $\mu$ -a.e. on $\Omega$ to a function $f \in L^{1}(\Omega , \mathcal{F}, \mu )$ and $\int f=\sum_{n=1}^{\infty } \int f_{n}$ .

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Problem 6.4.10

(a)(b)

Let $(\Omega , \mathcal{F}, \mu )$ denote a generic measure space, and let $(\Omega , \bar{\mathcal{F}}, \bar{\mu })$ denote the completion of $(\Omega , \mathcal{F}, \mu )$ . Let $\bar{f}$ denote an $\bar{\mathcal{F}}/\mathcal{B}^{*}$ -measurable function on $\Omega$ . [The bar on $f$ is just part of the notation; it denotes no operation.]

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(a)

There exists $f$ on $\Omega$ that is $\mathcal{F}/\mathcal{B}^{*}$ -measurable with $f=\bar{f}$ $\bar{\mu }$ -a.e. [Here, look back at the outline of Exercise 2 in Section 5.5, where there $\mathcal{F}$ consisted of the Borel sets and $\bar{\mathcal{F}}$ consisted of the Lebesgue sets. Observe that nothing specific to Euclidean space was used in that outline --- only the general results concerning completion --- hence the same outline will work here.]

(b)

If one of $\int \bar{f} \, d\bar{\mu }$ and $\int f \, d\mu$ exists, then so does the other and both integrals coincide. Therefore, integrals of $\bar{\mathcal{F}}/\mathcal{B}^{*}$ -measurable functions with respect to $\bar{\mu }$ are completely determined by the integrals of $\mathcal{F}/\mathcal{B}^{*}$ -measurable functions with respect to $\mu$ . [It is perhaps best to use the usual approach in proving results about integrals: show that the claim holds for simple, then nonnegative, then measurable functions.]