7

Since $fg\geq 1$, you have $\sqrt{f(x)\cdot g(x)} \geq 1$ for all $x$, and so $$ 1 \leq \int_X \sqrt{f}\cdot \sqrt{g} d\mu \leq \sqrt{\int_X f d\mu }\cdot \sqrt{\int_X g d\mu } $$ where the second inequality is Cauchy–Schwarz; squaring both sides gives the inequality.


6

It is sufficient to show that $m(E \cap [a, b]) = 0$ for every $a$ and $b$ with $a < b$. Let $t = \inf_{x \in [a, b]}{e^{-x^2/2}}$. Then $t > 0$ and we have $$0 = \int_{E}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}tdx = m(E \cap [a, b])t$$ and as $t > 0$, this implies $m(E \cap [a, b]) = 0$.


5

A measure space must specify three things: An underlying set; A $\sigma$-algebra of subsets which are measurable in this measure space; A measure (a function on this $\sigma$-algebra). When we say "a measurable subset of the real numbers" we typically mean "a subset of the real numbers measurable with respect to the Lebesgue measure", ...


5

The questions, despite looking as a representation problem in functional analysis, are much deeper as they bring out the history of the topic involved, notably $BV$-functions and the reasons why the customary definition adopted for the variation of a multivariate function is definition 2 above. And as thus the answers below needs to indulge a bit on this ...


5

$\bigcup_{i\ge 1}A_i$ is by definition simply $$\left\{x:\exists i\in\Bbb Z^+\,(x\in A_i)\right\}\,,$$ which clearly does not depend on the order in which the sets are enumerated. You can simply let $\mathscr{A}=\{A_i:i\in\Bbb Z^+\}$; then $$\bigcup_{i\ge 1}A_i=\bigcup\mathscr{A}=\{x:\exists A\in\mathscr{A}\,(x\in A)\}\,,$$ with no reference to the ...


5

Hint: Consider the sets $[-r,r] \subseteq \mathbb{R}$. Can you show $r \mapsto \mu(E \cup [-r,r] \cap F)$ is continuous? Can you show for any $r$ the set we're measuring is compact? What does the intermediate value theorem buy us? I hope this helps ^_^


5

This is false. For an easy counterexample, let $X_n\geq 0$ be any sequence of random variables with $\liminf_n X_n = 0$ which converges to $X = 1$ in probability. Then let $\mathcal{Q}$ be large enough so that the $X_n$ are $\mathcal{Q}$-measurable. The inequality in question then becomes $X \leq \liminf_n X_n,$ which is false almost surely. For instance, ...


5

It's true that the definition of absolute continuity is the same either we state it in the form of the finite and disjoint intervals or we state it in the form of countable and disjoint intervals. In detail, suppose we have the following definitions of absolute continuity: Definition 1 Let $f:\mathbb{R}\to \mathbb{R}$ be a function. We say that $f$ is ...


4

By assumption, as $Z \notin L^{\infty}$, we have $$\mathbb{P}(Z > n^2) > 0, \quad n \in \mathbb{N}.$$ Therefore, we may define $$X = \sum_{n=1}^{\infty}\frac{1}{n^2} \frac{\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)} \in L^1(\mathbb{P}).$$ However, it follows that \begin{align*}\mathbb{E}[ZX] = \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{1}{n^2} \frac{...


4

Cannot be true in general: consider $k=1$ (or any odd $k$), take $X_i$'s to be i.i.d normal with mean $0$ and variance $1$, we show below that $Y_n$ is not UI. Note that $$ \mathbb{E}\left(|Y_n| 1\{|Y_n|\geq M\}\right) \geq M\mathbb{P}\left(|Y_n|\geq M\right) = M\mathbb{P}\left(|\bar{X}_n|\leq \frac{1}{M}\right) \to M, $$ as $n\to \infty$ by laws of large ...


4

The assertion “not every subset of $\Bbb R$ is measurable” is valid for the Lebesgue measure. Anyway, here Axler is talking about $\sigma$-algebras, not about measures. And, yes, $\mathcal P(\Bbb R)$ is a $\sigma$-algebra. And, on this $\sigma$-algebra you can define the measure $m(X)=\#X$; with respect to this measure, $\Bbb R$ is measurable.


4

If $p > p_0$ and $f \in L^{p_0} \cap L^\infty$ then $|f|^p = |f|^{p - p_0}|f|^{p_0} \le \|f\|_\infty^{p - p_0}|f|^{p_0}$ almost everywhere.


4

The claim is true: Let $S = \sum_{i=1}^\infty \chi_{E_i}$. Suppose for the sake of contradiction that $S < \infty$ a.e. Then, there exists an $M$ such that $\Pr(S\le M) \ge 1 - \frac\varepsilon 2$. If $A$ is the event that $S\le M$, then $\mu(A\cap E_i) \ge \frac\varepsilon 2$ for all $i$. But this would imply $\int_A S\,d\mu = \sum_{i=1}^\infty \mu(A\...


4

Well, $[0,1]=C\hspace{1mm}\cup C^{c} $ (I discourage you to use $\mathbb{C}$ for the cantor set) and you can write f as: $$f(x)=0\cdot\chi_{C}(x)+x^{2}\chi_{C^{c}}$$ where by $\chi$ I mean the characteristic function of the set in subscript. The sum of measurable functions is measurable as well as the product, and polynomial functions are measurable.


4

The family of measurable sets form a $\sigma$-algebra but not all sets belonging in any $\sigma$-algebra are measurable. So to fully define a measure space we need a set $X$, a $\sigma$-algebra $\mathcal{A}$ on $X$ and a function $\mu: \mathcal{A} \to [0, \infty]$ which satisfies some properties. A set $A \subseteq X$ is called $\mu$-measurable (or simply ...


4

The symmetrization is called Khintchine's inequality, see here: https://en.wikipedia.org/wiki/Khintchine_inequality For the contraction principle, I don't know any reference, but I can prove it. First note that if $Z\geq 0$ and $p > 0$, then \begin{align} \mathbb{E}[Z^p] = \mathbb{E}\Big[\int_0^{Z} p t^{p-1} dt\Big] = \mathbb{E}\Big[\int_0^{\infty} p t^{p-...


4

A set $\mathfrak B$ satisfies the definition in your book if and only if $\mathfrak B$ is a $\sigma$-algebra (in the wikipedia sense) on $\cup\mathfrak B$: for the noteworthy algebraic detail, notice that $A\setminus B=A\triangle (A\cap B)$. This is also equivalent to the property of $\mathfrak B$ being a $\sigma$-algebra in the wikipedia sense over some set....


4

Let me discuss the case $n=1$ only. Let me denote by $f_x$ the weak derivative of $f$ with respect to $x$. Let $\phi,\psi$ be two smooth test functions. Then $$ -\iint f_x(x,y) \phi(x)\psi(y)dx\ dy=\iint f(x,y) \phi_x(x)\psi(y)dx\ dy, $$ using Fubini on both sides gives $$ \int \left( \int -f_x(x,y) \phi(x) + f(x,y) \phi_x(x)\ dx\right) \psi(y) \ dy = 0 $$ ...


3

You prove that by giving an example of a set of subsets that is closed under intersections and differences but not closed under unions. E.g. on any set $X$ with at least two points, the family $\{\{x\}: x \in X\} \cup \{\emptyset\}$ is such. so being closed under union and differences (a ring) implies being closed under intersections, but being closed under ...


3

I have recently got interested in continued fractions, so I'll give it a try! CONTINUED FRACTIONS. It is a basic property of continued fractions that if $h_n/k_n$ is the sequence of convergent to $x\in\mathbb R\setminus\mathbb Q$, then $$ \frac 1{k_n(k_n+k_{n+1})}<\left|x-\frac{h_n}{k_n}\right|<\frac 1{k_nk_{n+1}}. $$ It follows that for all $h,k$ ...


3

The set $\mathcal{P}(X)$ of all subsets of $X$ is a sigma algebra because it satisfies the axioms for that kind of structure. That is independent of any discussion of measures. The definition of a measure space is a pair $(X,S)$ where $S$ is some subset of $\mathcal{P}(X)$ that happens to be a sigma algebra. It may or may not be all of $\mathcal{P}(X)$. ...


3

Suppose that such measure exists. Then taking countable cover by open sets of diameter $\leq 1$, one of those sets, say $U_1$, must have measure $1$. Cover $U_1$ by countable amount of open sets of diameter $\leq 1/2$ and take the one having measure $1$. This process gives us a decreasing sequence of open sets $U_k$ with diameter $\leq 1/k$ and $\mu(U_k) = 1$...


3

This works. Your argument shows that if $f$ is bounded on a set $E$ of Lebesgue measure zero, then $\int_E f = 0$. In fact the stronger claim that $\int_E f = 0$ without any hypothesis on $f$ holds! Indeed the integral is, by definition, the supremum over $g$ with $g$ simple and $g \leq f$ of $\int_E g$. But any such $g$ has integral zero.


3

Welcome to MSE! Hint: It's often useful to split up a random variable into the places where it is "good" and "bad". Then we can control those regions separately to get whatever inequality we're interested in. For this, we might try looking at the decomposition $$ X = X \cdot \mathbf{1}_{\{X \leq \lambda a\}} + X \cdot \mathbf{1}_{\{X >...


3

According to the p-Adics section of the reference manual there is nothing I could find to do with zeta functions nor integration. At any rate, here's a suggestion on algorithmically getting an arbitrary level of accuracy on your integral. In order to be called a p-adic measure we require it to be a bounded p-adic distribution, and this is enough to imply ...


3

By definition, $\int f$ is the supremum of $\int h$ for $0 \leq h \leq f$ bounded measurable of finite support. So at the very end, you just need to take the sup over such $h$ to get $\int f \leq \liminf \int f_n$. He uses such $h$ is order to apply the bounded dominated convergence theorem, which he proves earlier. This allows him to prove Fatou in terms of ...


3

You can also let $g_n=inf_{i \geq n} \{f_n\}$ and note that $\int g_n \leq \int f_n$ for each n. Moreover $g_n$ increases monotonically to lim inf $f_n$ . Hence by monotone convergence theorem $\int g_n = \int lim inf f_n$. Since $\int g_n \leq \int f_n$ we have $\int lim inf f_n \leq lim inf \int f_n$. The final step is to note that pointwise convergence ...


3

If you are familiar with dominated convergence notice that $f_n(X)\xrightarrow{n\rightarrow\infty} X$ pointwise, $|f_n(X)|\leq |X|$, $X$ is assumed to be integrable ($E[|X|]=\int |X|\,dP<\infty$). Then by dominated convergence $$ \int f_n(X)\,dP\xrightarrow{n\rightarrow\infty}\int X\,dP=E[X]$$ similarly for $Y$. There is still the issue of the ...


3

Let $f$ be the limit of $\chi_{E_n}$ in $L^p$. All you have to do is show that $f(x) = 0$ or $f(x) = 1$ almost everywhere. Then we see that $f = \chi_E$ almost everywhere, where $E = \{x \in X | f(x) = 1\}$ To do this, it is helpful to define $g(x) = \min(|x - 1|, |x|)$. We will show that $g \circ f = 0$ almost everywhere. We first show that $\int g(f(x))^p \...


3

Let $f \geq 0$ and let $M=\sup \{f(x): x\in X\}$. Then $\|M-f\|\leq M$ since $f(x) \in [0,M]$ for all $x$. If $\phi(f)$ is real we can finish the proof as follows: $M-\phi (f)=\phi(M-f) \leq \|\phi\| \|M-f||\leq M$. Thus $\phi (f) \geq 0$. To show that $\phi (f)$ is real consider following: $|\phi (f) \pm in|=|\phi(f \pm in)|\leq \|f\pm in\|\leq \sqrt {M^{...


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