## Exercises in Real Analysis II

This post is a sequel to Exercises in Real Analysis.

4. Let $T$ be the triangle $\{(x, y) \in {\mathbb{R}}^2: 0 \leq |x| \leq y \leq 1\}$, and $\mu$ be the restriction of the planar Lebesgue measure on $T$. Suppose that $f \in L^2(T, \mu)$. Prove that

$\liminf_{y \rightarrow 0^+} \int_{-y}^y |f(x, y)| dx = 0$.

Solution. Assume on the contrary that

$L = \liminf_{y \rightarrow 0^+} \int_{-y}^y |f(x, y)| dx > 0$.

Then for some $\delta > 0$, we have

$\int_{-y}^y |f(x, y)| dx > \frac{L}{2}$      (1)

for all $0 < y < \delta$.

To link $L^1$ to $L^2$, we use the Cauchy-Schwarz inequality:

$\int_{-y}^y |f(x, y)| dx \leq \left( \int_{-y}^y |f(x, y)|^2 dx \right)^{\frac{1}{2}} (2y)^{\frac{1}{2}}$.

Hence from (1), we get

$\int_{-y}^y |f(x, y)|^2 dx \geq \frac{L^2}{8} \frac{1}{y}$,    $0 < y < \delta$.

Integrating over $y$ (Fubini’s theorem), we get

$\|f\|^2_{L^2(T, \mu)} \geq \int_0^{\delta} \int_{-y}^y |f(x, y)|^2 dx dy \geq \int_0^{\delta} \frac{L^2}{8} \frac{1}{y} dy = \infty$.

Contradiction. $\Box$

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5. Let $X$ be a compact metric space, and $\mu$ be a finite positive Borel measure on $X$. Suppose that $\mu(\{x\}) = 0$ for every $x \in X$. Prove that for every $\epsilon > 0$ there is a $\delta > 0$ such that, if $E$ is any Borel subset of $X$ having diameter less than $\delta$, then $\mu(E) < \epsilon$.

Solution. Since $\mu(X) < \infty$, by continuity of measure we have

$\lim_{r \downarrow 0} \mu(B(x, r)) = 0$

for any $x \in X$. Thus, given $\epsilon > 0$, there exists $r_x > 0$ such that

$\mu(B(x, r_x)) < \epsilon$.

Clearly $\{B(x, r_x)\}_{x \in x}$ is an open cover of $X$, and by compactness there is a finite subcover, say

$X \subset B(x_1, r_{x_1}) \cup \cdots \cup B(x_n, r_{x_n})$.

Recall that the finite cover has some Lebesgue number $\delta > 0$, i.e., every subset $E$ of $X$ with diameter less than $\delta$ is contained entirely in some $B(x_i, r_{x_i})$. Hence (assuming $B$ is Borel)

$\mu(E) \leq \mu(B(x_i, r_{x_i})) < \epsilon$.   $\Box$

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6. Let $X$ and $Y$ be compact metric spaces, and let $C(X)$ and $C(Y)$ denote the Banach spaces of continuous, real-valued functions on $X$ and $Y$ respectively (with sup norm). Suppose that $\phi: X \rightarrow Y$ is a continuous surjective map. Let

$D = \{f \in C(X): f(x) = f(x') \text{ whenever } \phi(x) = \phi(x')\}$.

(a) Show that $D$ is a closed subspace of $C(X)$, and that

$D = \{g \circ \phi: g \in C(Y)\}$.

(b) Let $\nu$ be a finite positive Borel measure on $Y$. Prove that there is a finite positive Borel measure $\mu$ on $X$ such that $\mu(\phi^{-1}(F)) = \nu(F)$ for all Borel subsets $F$ of $Y$.

Solution. (a) It is easy to show that $D$ is a closed subspace.

It is also clear that $\{g \circ \phi: g \in C(Y)\} \subset D$. To show the reverse inclusion, consider any $f \in D$. Since $\phi$ is surjective, for each $y \in Y$ there is some $x \in X$ such that $\phi(x) = y$. We define a function $g$ on $Y$ by setting $g(y) = f(x)$. The function $g$ is well defined, for if $\phi(x) = \phi(x') = y$, then $f(x) = f(x')$ since $f \in D$.

It remains to show that $g$ is continuous. Fix $y \in Y$ and suppose $y_n \in Y \rightarrow y$. Fix $x_n \in X$ such that $\phi(x_n) = y_n$. Then

$g(y_n) = f(x_n)$.

Since $X$ is compact, some subsequence $x_{n_k}$ converges to some $x \in X$. By continuity of $\phi$,

$\phi (x) = \lim_n \phi (x_n) = \lim_n y_n = y$.

Finally, by continuity of $f$, we have

$g(y) = f(x) = \lim_n f(x_n) = \lim_n g(y_n)$.

Hence $f \in \{g \circ \phi: g \in C(Y)\}$.

(b) To construct a measure on a compact metric space, a good idea is to use the Riesz representation theorem.

Define a positive linear functional $\Phi$ on the subspace $D$ by setting

$\Phi f = \int_Y g(y) d\nu(y)$    if $f = g \circ \phi \in D$.

(You should verify that this definition makes sense.) The linear functional $\Phi$ is bounded on $D$, since

$|\Phi f| \leq \|g\|_{sup} \nu(X) = \|f\|_{sup} \nu(X)$.

(You should verify the last inequality.) To use the Riesz representation theorem, we need a linear functional defined on the whole $C(X)$. To get such a functional, we simply use the Hahn-Banach theorem to extend $\Phi$ to a bounded linear functional on $C(X)$. (Note that the question only requires us to construct a measure. Uniqueness is not guarenteed.)

Now, the Riesz representation theorem provides us with a positive Borel measure $\mu$ on $X$ such that

$\Phi f = \int_X f(x) d\mu(x)$,    $f \in C(X)$.

And when $f = g \circ \phi \in D$, we have

$\Phi f = \int_X g \circ \phi (x) d\mu(x) = \int_X g(y) d\nu(y)$.   (2)

Finally, we want to show that (2) implies that

$\mu(\phi^{-1}(F)) = \nu(F)$   (3)

for all Borel subsets $F$ of $Y$.

It suffices to prove (3) where $F$ is open. Fix such a set $F$. Then, there exists a sequence $g_n(x) \in C(X)$ such that $g_n \uparrow \chi_F$

$\int_X g_n \circ \phi (x) d\mu(x) = \int_X g_n(y) d\nu(y)$.

By the monotone convergence theorem, we get

$\mu(\phi^{-1}(F)) = \int_X \chi_F \circ \phi(x) d\mu(x) = \int_X \chi_F(y) d\nu(y) = \nu(F)$. $\Box$

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### 2 Responses to Exercises in Real Analysis II

1. KKK says:

Here is another solution for Real analysis prelim provided by Wongting.

Question:
(a) For a nonnegative ${L^1}$ function ${f}$ on ${[0, \infty)}$, show that there is a nonnegative strictly increasing function ${a(x)}$ with ${\displaystyle\lim_{x\rightarrow \infty}a(x)=\infty}$ and ${af\in L^1}$.
(b) For a sequence ${f_n}$ of nonnegative ${L^1}$ functions on ${[0, \infty)}$ with ${|f_n|_{L^1}\rightarrow 0}$, find a nonnegative strictly increasing function ${b(x)}$ with ${\displaystyle\lim_{x\rightarrow \infty}b(x)=\infty}$ and ${|bf_n|_{L^1}\rightarrow0}$.

Solution:
(a) Let ${I=\int_0^\infty f}$. We can choose ${0=x_0 with ${x_i\rightarrow\infty}$ such that

$\displaystyle \int_{x_i}^\infty f \leq \frac{I}{2^i}.$

We then define ${a(x_i)=i}$ and is linear between ${x_i}$‘s. Clearly we only have to check ${af\in L^1}$.

To see this, let $t_0=0$ and ${t_i= \inf\{t\geq 0: \int_0^t f=\frac{I}{2^1}+\cdots +\frac{I}{2^i}\}}$, $i\geq 1$. Observe that ${t_i\leq x_i}$, and so ${a(t_i)\leq i}$. We then have

$\displaystyle \int_0^\infty af= \sum_{i=1}^\infty \int_{t_{i-1}}^{t_{i}}af\leq \sum _{i=1}^\infty \int_{t_{i-1}}^{t_{i}}i f= \sum _{i=1}^\infty \frac{iI}{2^i}=CI$

where ${\displaystyle C=\sum _{i=1}^\infty \frac{i}{2^i}<\infty}$. Thus ${af\in L^1}$.
(b) We modify the above proof. Let ${I_n:=\int _0^\infty f_n}$. There is ${x_1}$ such that

$\displaystyle \int_{x_1}^\infty f_1 \leq\frac{I_1}{2}.$

Inductively, we can find ${0=x_0 with ${x_i\rightarrow \infty }$ such that for all ${n\leq i}$,

$\displaystyle \int_{x_i}^\infty f_n \leq\frac{I_n}{2^i}.$

We then choose ${b(x)}$ exactly as before, i.e. ${b(x_i)=i}$ and linear in between. The argument in (a) then shows that ${\int bf_n\leq CI_n\rightarrow 0}$, noting that ${C}$ is independent of ${n}$.

2. Leonard Wong says:

Nice proof!

In the last sentence, it should be $\int bf_n \leq C I_n$. [Yes, thanks! -KKK]