Math Problems Vault

Problem 15.1

Suppose that $f$ is measurable and nonnegative. Show that $\int^{*} f d \mu =\infty$ if $\mu [\omega$ : $f(\omega )>0]=\infty$ or if $\mu [\omega : f(\omega )>a]>0$ for all $a$

There are many functions familiar from calculus that ought to be integrable but are of the types in the preceding problem and hence have infinite upper integral. Examples are $x^{-2} I_{(1, \infty )}(x)$ and $x^{-1 / 2} I_{(0,1)}(x)$ . Therefore, (15.10) is inappropriate as a definition of $\int f d \mu$ for nonnegative $f$ . The only problem with (15.10), however, is that it treats infinity the wrong way. To see this, and to focus on essentials, assume that $\mu (\Omega )<\infty$ and that $f$ is bounded, although not necessarily nonnegative or measurable $\mathscr {F}$ .

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Problem 15.2

(a)(b)

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(a)

Show that

\sum _{i}\left[\inf _{\omega \in A_{i}} f(\omega )\right] \mu \left(A_{i}\right) \leq \sum _{j}\left[\inf _{\omega \in B_{i}} f(\omega )\right] \mu \left(B_{j}\right)

if $\left\{ B_{j}\right\}$ refines $\left(A_{i}\right)$ . Prove a dual relation for the sums in (15.10) and conclude that

\int _{*} f d \mu \leq \int ^{*} f d \mu \tag {15.11}

(b)

Now assume that $f$ is measurable $\mathscr {F}$ and let $M$ be a bound for $\left|f\right|$ . Consider the partition $A_{i}=[\omega : i \epsilon <f(\omega ) \leq (i+1) \epsilon ]$ , where $i$ ranges from $-N$ to $N$ and $N$ is large enough that $N \epsilon >M$ . Show that

\sum _{i}\left[\sup _{\omega \in A_{i}} f(\omega )\right] \mu \left(A_{i}\right)-\sum _{i}\left[\inf _{\omega \in A_{i}} f(\omega )\right] \mu \left(A_{i}\right) \leq \epsilon \mu (\Omega ).

Conclude that

\int _{*} f d \mu =\int ^{*} f d \mu \tag {15.12}

To define the integral as the common value in (15.12) is the Darboux-Young approach. The advantage of (15.3) as a definition is that (in the nonnegative case) it applies at once to unbounded $f$ and infinite $\mu$

Problem 15.3

3.2 15.2 ↑ For $A \subset \Omega$ , define $\mu^{*}(A)$ and $\mu_{*}(A)$ by (3.9) and (3.10) with $\mu$ in place of $P$ . Show that $\int^{*} I_{A} d \mu =\mu^{*}(A)$ and $\int_{*} I_{A} d \mu =\mu_{*}(A)$ for every $A$ . Therefore, (15.12) can fail if $f$ is not measurable $\mathscr {F}$ . (Where was measurability used in the proof of (15.12)?)

The definitions (15.3) and (15.6) always make formal sense (for finite $\mu (\Omega )$ and supif D, but they are reasonable-accord with intuition-only if (15.12) holds. Under what conditions does it hold?

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Problem 15.4

(a)(b)

$10.515 .3 \uparrow$

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(a)

Suppose of $f$ that there exist an F-set $A$ and a function $g$ , measurable $\mathscr {F}$ , such that $\mu (A)=0$ and $[f \neq g] \subset A$ . This is the same thing as assuming that $\mu^{*}[f \neq g]=0$ , or assuming that $f$ is measurable with respect to $\mathscr {F}$ completed with respect to $\mu$ . Show that (15.12) holds.

(b)

Show that if (15.12) holds, then so does the italicized condition in part (a).

Rather than assume that $f$ is measurable $\mathscr {F}$ , one can assume that it satisfies the italicized condition in Problem 15.4(a)-which in case $(\Omega , \mathscr {F}, \mu )$ is complete is the same thing anyway. For the next three problems, assume that $\mu (\Omega )<\infty$ and that $f$ is measurable $\mathscr {F}$ and bounded.

Problem 15.5

↑ Show that for positive $\epsilon$ there exists a finite partition $\left\{ A_{i}\right\}$ such that, if $\left\{ B_{j}\right\}$ is any finer partition and $\omega_{j} \in B_{j}$ , then

\left|\int f d \mu -\sum _{j} f\left(\omega _{j}\right) \mu \left(B_{j}\right)\right|<\epsilon .

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Problem 15.6

↑ Show that

\int f d \mu =\lim _{n} \sum _{\left|k\right| \leq n 2^{n}} \frac{k-1}{2^{n}} \mu \left[\omega : \frac{k-1}{2^{n}} \leq f(\omega )<\frac{k}{2^{n}}\right].

The limit on the right here is Lebesgue's definition of the integral.

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Problem 15.7

↑ Suppose that the integral is defined for simple nonnegative functions by $\int \left(\sum_{i} x_{i} I_{A}\right) d \mu =\sum_{i} x_{i} \mu \left(A_{i}\right)$ . Suppose that $f_{n}$ and $g_{n}$ are simple and nondecreasing and have a common limit: $0 \leq f_{n} \uparrow f$ and $0 \leq g_{n} \uparrow f$ . Adapt the arguments used to prove Theorem 15.1(iii) and show that $\lim_{n} \int f_{n} d \mu =\lim_{n} \int g_{n} d \mu$ . Thus, in the nonnegative case, $\int f d \mu$ can (Theorem 13.5) consistently be defined as $\lim_{n} \int f_{n} d \mu$ for simple functions for which $0 \leq f_{n} \uparrow f$ .

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