Math Problems Vault

Problem 3B.1

Give an example of a sequence $f_{1}, f_{2}, \ldots$ of functions from $\mathbb {N}$ to $[0, \infty )$ such that

\lim _{k \rightarrow \infty } f_{k}(m) = 0

for every $m \in \mathbb {N}$ but $\lim_{k \rightarrow \infty } \int f_{k} d \mu = 1$ , where $\mu$ is counting measure on $\mathbb {N}$ .

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Problem 3B.2

Give an example of a sequence $f_{1}, f_{2}, \ldots$ of continuous functions from $\mathbb {R}$ to $[0,1]$ such that

\lim _{k \rightarrow \infty } f_{k}(x) = 0

for every $x \in \mathbb {R}$ but $\lim_{k \rightarrow \infty } \int f_{k} d \lambda = \infty$ , where $\lambda$ is Lebesgue measure on $\mathbb {R}$ .

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Problem 3B.3

Suppose $\lambda$ is Lebesgue measure on $\mathbb {R}$ and $f: \mathbb {R} \rightarrow \mathbb {R}$ is a Borel measurable function such that $\int \left|f\right| d \lambda < \infty$ . Define $g: \mathbb {R} \rightarrow \mathbb {R}$ by

g(x) = \int _{(-\infty , x)} f d \lambda

Prove that $g$ is uniformly continuous on $\mathbb {R}$ .

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Problem 3B.4

(a)(b)(c)

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(a)

Suppose $(X, \mathcal{S}, \mu )$ is a measure space with $\mu (X) < \infty$ . Suppose that $f: X \rightarrow [0, \infty )$ is a bounded $\mathcal{S}$ -measurable function. Prove that

\int f d \mu = \inf \left\{ \sum _{j = 1}^{m} \mu \left(A_{j}\right) \sup _{A_{j}} f: A_{1}, \ldots , A_{m} \text{ is an } \mathcal{S} \text{-partition of } X\right\}

(b)

Show that the conclusion of part (a) can fail if the hypothesis that $f$ is bounded is replaced by the hypothesis that $\int f d \mu < \infty$ .

(c)

Show that the conclusion of part (a) can fail if the condition that $\mu (X) < \infty$ is deleted.

Remark.

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Part (a) of this exercise shows that if we had defined an upper Lebesgue sum, then it could be used to define $\int f d \mu$ when $f$ is bounded and $\mu (X) < \infty$ . However, parts (b) and (c) show that the hypotheses that $f$ is bounded and that $\mu (X) < \infty$ are needed if defining the integral via the equation above. The definition of the integral via the lower Lebesgue sum does not require these hypotheses, showing the advantage of using the lower Lebesgue sum.

Problem 3B.5

Let $\lambda$ denote Lebesgue measure on $\mathbb {R}$ . Suppose $f: \mathbb {R} \rightarrow \mathbb {R}$ is a Borel measurable function such that $\int \left|f\right| d \lambda < \infty$ . Prove that

\lim _{k \rightarrow \infty } \int _{[-k, k]} f d \lambda = \int f d \lambda

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Problem 3B.6

Let $\lambda$ denote Lebesgue measure on $\mathbb {R}$ . Give an example of a continuous function $f:[0, \infty ) \rightarrow \mathbb {R}$ such that $\lim_{t \rightarrow \infty } \int_{[0, t]} f d \lambda$ exists (in $\mathbb {R}$ ) but $\int_{[0, \infty )} f d \lambda$ is not defined.

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Problem 3B.7

Let $\lambda$ denote Lebesgue measure on $\mathbb {R}$ . Give an example of a continuous function $f:(0,1) \rightarrow \mathbb {R}$ such that $\lim_{n \rightarrow \infty } \int_{\left(\frac{1}{n}, 1\right)} f d \lambda$ exists (in $\mathbb {R}$ ) but $\int_{(0,1)} f d \lambda$ is not defined.

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Problem 3B.8

Verify the assertion in 3.38.

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Remark.

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Setup: suppose $a < b$ and $f:[a, b] \rightarrow \mathbb {R}$ is a bounded function. For $n \in \mathbb {N}$ , let $P_{n}$ be the partition that divides $[a, b]$ into $2^{n}$ subintervals of equal size, with corresponding closed subintervals $I_{1}, \ldots , I_{2^{n}}$ , and let

g_{n} = \sum _{j = 1}^{2^{n}}\left(\inf _{I_{j}} f\right)\chi _{I_{j}} \qquad \text{and} \qquad h_{n} = \sum _{j = 1}^{2^{n}}\left(\sup _{I_{j}} f\right)\chi _{I_{j}}.

Then $g_{1} \leq g_{2} \leq \cdots$ is an increasing sequence and $h_{1} \geq h_{2} \geq \cdots$ is a decreasing sequence of functions on $[a, b]$ . Define $f_{L}, f_{U}:[a, b] \rightarrow \mathbb {R}$ by

f_{L}(x) = \lim _{n \rightarrow \infty } g_{n}(x) \qquad \text{and} \qquad f_{U}(x) = \lim _{n \rightarrow \infty } h_{n}(x).

3.38: $\left\{ x \in [a, b]: f_{U}(x) \neq f_{L}(x)\right\} = \left\{ x \in [a, b]: f \text{ is not continuous at } x\right\}$ .

Problem 3B.9

Verify the assertion in Example 3.41.

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Remark.

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Example 3.41: if $f$ is a measurable function and $\int \left|f\right|^{p} d\mu < \infty$ for all $p > 0$ , it does not imply that $f$ is bounded.

Problem 3B.10

(a)(b)

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(a)

Suppose $(X, \mathcal{S}, \mu )$ is a measure space such that $\mu (X) < \infty$ . Suppose $p, r$ are positive numbers with $p < r$ . Prove that if $f: X \rightarrow [0, \infty )$ is an $\mathcal{S}$ -measurable function such that $\int f^{r} d \mu < \infty$ , then $\int f^{p} d \mu < \infty$ .

(b)

Give an example to show that the result in part (a) can be false without the hypothesis that $\mu (X) < \infty$ .

Problem 3B.11

Suppose $(X, \mathcal{S}, \mu )$ is a measure space and $f \in \mathcal{L}^{1}(\mu )$ . Prove that

\left\{ x \in X: f(x) \neq 0\right\}

is the countable union of sets with finite $\mu$ -measure.

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Problem 3B.12

Suppose

f_{k}(x) = \frac{(1-x)^{k} \cos \frac{k}{x}}{\sqrt{x}}

Prove that $\lim_{k \rightarrow \infty } \int_{0}^{1} f_{k} = 0$ .

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Problem 3B.13

Give an example of a sequence of nonnegative Borel measurable functions $f_{1}, f_{2}, \ldots$ on $[0,1]$ such that both the following conditions hold:

$\lim_{k \rightarrow \infty } \int_{0}^{1} f_{k} = 0$ ;
$\sup_{k \geq m} f_{k}(x) = \infty$ for every $m \in \mathbb {N}$ and every $x \in [0,1]$ .

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Problem 3B.14

(a)(b)(c)

Let $\lambda$ denote Lebesgue measure on $\mathbb {R}$ .

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(a)

Let $f(x) = 1 / \sqrt{x}$ . Prove that $\int_{[0,1]} f d \lambda = 2$ .

(b)

Let $f(x) = 1 /\left(1+x^{2}\right)$ . Prove that $\int_{\mathbb {R}} f d \lambda = \pi$ .

(c)

Let $f(x) = (\sin x) / x$ . Show that the integral $\int_{(0, \infty )} f d \lambda$ is not defined but $\lim_{t \rightarrow \infty } \int_{(0, t)} f d \lambda$ exists in $\mathbb {R}$ .

Problem 3B.15

Prove or give a counterexample: If $G$ is an open subset of $(0,1)$ , then $\chi_{G}$ is Riemann integrable on $[0,1]$ .

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Problem 3B.16

(a)(b)

Suppose $f \in \mathcal{L}^{1}(\mathbb {R})$ .

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(a)

For $t \in \mathbb {R}$ , define $f_{t}: \mathbb {R} \rightarrow \mathbb {R}$ by $f_{t}(x) = f(x-t)$ . Prove that

\lim _{t \rightarrow 0}\left\| f-f_{t}\right\| _{1} = 0

(b)

For $t > 0$ , define $f_{t}: \mathbb {R} \rightarrow \mathbb {R}$ by $f_{t}(x) = f(t x)$ . Prove that

\lim _{t \rightarrow 1}\left\| f-f_{t}\right\| _{1} = 0