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13

A standard argument to show that $\mathcal{L}(\mathbb R)$ has the same size as $\mathcal{P}(\mathbb R)$ is to note that the Cantor subset of $[0,1]$ has the same size as $\mathbb R$ and measure 0, so any of its subsets also has measure 0. You can use the same idea to find the size of $\mathcal P(\mathbb R)\setminus\mathcal L(\mathbb R)$: Fix a nonmeasurable ...


7

HINT: You can add $\Bbb Q\cap[0,1]$ to any measurable subset of $[0,1]$ without changing its measure.


5

Let me try to explain the Vitali construction by breaking everything into small pieces. (Usually I find this presented quite rapidly in books, so that everything seems to come all at once and it's not clear what all the pieces even were.) You begin by quotienting $[0,1]$ by this equivalence relation that you stated. You get the full collection of ...


5

Another idea: Put $g=\sum |f_n|$. We see that (as all the functions are $\geq 0$) we have $$\int_0^1g(t)dt=\sum _{n\geq 1}\int_0^1|f_n(t)|dt\leq \sum_{n\geq 1}\frac{1}{n^2}<+\infty$$ Hence $g$ is in $L^1$, hence finite $a.e$, and the serie $|f_n|$ is a.e convergent, and $f_n\to 0$ a.e.


3

A similar limit is used to prove Euler's limit product formula for the $\Gamma$ function: $$ t! = \lim_{n\to +\infty}\frac{n! n^t}{(t+1)(t+2)\cdot\ldots\cdot(t+n)}\tag{1}.$$ For any $x\in\mathbb{R}^+$ we have $$ \max_{n\geq x} \left(1-x/n\right)^n\leq e^{-x}\tag{2} $$ and $e^{-x}\cdot e^{x/2}\in L^1(\mathbb{R}^+)$, hence by monotone or dominated convergence ...


3

Let $\{E_n\}$ be a partition of a measurable set $E$ (i.e. a countable mutually disjoint collection whose union is $E$). then for $m\in\mathbb{N}$, the collection $\{E_1,\ldots,E_m,\bigcup_{n=m+1}^\infty E_n\}$ is a finite partition of $E$, and thus $$\sum_{n=1}^m|\mu(E_n)|+|\mu\left(\bigcup_{n=m+1}^\infty E_n\right)|\leq \mu_A(E) $$ Since $m$ was ...


3

Yes, it does follow. Notice that $x \in \lim \inf_{i \to \infty} A_i$ iff there is some $j$ such that $x \in A_{i}$ for all $j \le i$. In this case we clearly have $\lim \inf_{i \to \infty} \chi_{A_i} (x) = 1$. Conversely, if $\lim \inf_{i \to \infty} \chi_{A_i}(x) = 1$, then there is no infinite subsequence $(A_{i_k})_{k}$ such that $x \not \in A_{i_k}$ for ...


3

Your instinct is on track. The "substitution rule" is related to a sort of change of measure, that of "image measure". Let's suppose that $g:I\to J$ is a strictly increasing $C^1$ of $I$ onto $J$, where $I$ and $J$ are closed intervals contained in $\Bbb R$. Let $f:J\to\Bbb R$ be integrable, and define a measure $\mu$ by the formula $\mu(B)=\int_B f(u)\,du$ ...


3

I don't know how familiar with Banach spaces you are. It's a standard fact that "absolutely summable series converge" in a Banach space (in the same way this is true for series of numbers). More precisely, if $(v_{n})_{n \in \mathbb{N}}$ is a sequence in a Banach space $B$ and $\sum_{n = 1}^{\infty} \|v_{n}\|_{B} < \infty$, then $\sum_{n =1}^{N} v_{n}$ ...


2

Let $\epsilon>0$ and $\mathbb{R}=\cup_n (-n,n)=\cup_n I_n$. Now, by Egorov, for each $k$ exist $E_k^n$ such that $|I_n-E_k^n|<\epsilon/k$ Note that for each n, $E_k^n$ can be increasing and then $I_n-E_k^n$ is decreasing in k $|\mathbb{R}-\cup_{n,k}E_k^n|=|(\cup_n(\cap_k( I_n-E_k^n))|=|\cup_n(\cap_k(I_n-E_k^n))|\leq \sum_n(\lim_k |I_n-E_k^n|)=0$.


2

$x \in \liminf_i A_i$ iff $\exists j_0$ such that $x \in \bigcap_{i=j_0}^{\infty} A_i$ (by the definition of a union). So in total we know that $$x \in \liminf_i A_i \text{ iff }\, \exists j_0 : \forall i \ge j_0 : x \in A_i$$ which exactly says that $x$ lies in all but finitely many of the $A_i$: the only possible exceptions are the $A_i$ with $i < ...


2

Not in general. First, note that if $XY,Z$ are independent: $$ \mathbb{E}[XY\mid Z] = \mathbb{E}[XY] $$ which is "just a number." Now, it's tempting to try and say something in the case that if $XY,Z$ are not independent. But... If $X=f(Z)$ for some function $f$, then: $$ \mathbb{E}[XY\mid Z] = \mathbb{E}[f(Z)Y\mid Z] = f(Z)\mathbb{E}[Y\mid Z] = X\mathbb{...


2

This is a standard kind of proof that is used in situations where we have inductively defined collections. In this case, the inductively defined collection is the class of Borel sets. The Borel sets on a space are the smallest class of subsets of that space which contains all the open sets and is closed under countable unions and under complements. The ...


2

First question is yes. Given $|f|\leq M$ is bounded, we can approximate it uniformly using simple functions by cutting $[-M, M]$ into finitely many intervals $I_n$ with length $\epsilon$ and defining $\phi_n (x) =\sum_n \inf I_n \chi_{f^{-1}(I_n)}(x) $ . We would have $\|f - \phi_n\|_\infty \leq \epsilon$. $$\lim_m\int f d\mu_m = \lim_m \lim_n \int \phi_n ...


2

The statement is true if $A+B$ is measurable, this is the critical part in the assumption. If $A,B$ are compact, then so is $A+B$ and therefore measureable. Here is a proof for the theorem with the assumption of $A+B$ to be measurable using the statement for compact $A,B$: For arbitrary $\tilde A,\tilde B$ compact with $\tilde A\subset A$ and $\tilde B\...


2

Yes, that's acceptable. I would say: let $\Sigma$ be a σ-algebra. Then $\Sigma$ satisfies the first two semiring properties because, respectively, $\Sigma$ contains the empty set and $\Sigma$ is closed under finite intersections by virtue of being a σ-algebra. For the third property, fix $S, T \in \Sigma$ and let $U$ be the difference $U=S - T \...


2

Use the dominated convergence theorem with $\max(1,|f(x)|^2)$ as the dominating function.


2

Hint: $$\left\{x\in R |f(x)=g(x)\right\}=\left\{x\in R |f(x)-g(x)=0\right\}=(f-g)^{-1}\{0\}.$$ Questions for you: Is the singleton $\{ 0\}$ measurable? Is the sum/difference of measurable function measurable? Once you have established the above, the claim follows by very definition of measurability.


2

Note that a sequence of real numbers converges to a limit $L$ iff every subsequence has a subsubsequence that converges to $L$. Since your hypotheses are preserved by passing to subsequences, it suffices to show that $\int|f-f_{n_k}|\to 0$ for some subsequence $(f_{n_k})$. In particular, since $\int g_n\to 0$, there is a subsequence $(g_{n_k})$ of $(g_n)$ ...


2

You want sets that are independent in the sense of probability. Simple example: $$A_1=[0,1/2],$$ $$A_2=[0,1/4]\cup[1/2,3/4],$$ $$A_3=[0,1/8]\cup[1/4,3/8]\,\cup[1/2,5/8]\cup[3/4,7/8],$$ $$\dots$$


1

The lebegue measure is invariant by translations so the answer is yes it is. Just writing the definition also the second follow.


1

As Prahlad Vaidyanathan mentioned in the comments, consider the delta measure. Fix $x \in X$ and let \begin{align*} m_x(A) = \begin{cases} 1 & x \in A \\ 0 & x\not\in A \end{cases}. \end{align*} Then $m_x$ is a finite measure on $X$. This can be generalized to $m = \sum_{x \in X} p_x \mu_x$ with $p_x \in [0, \infty]$. Another possible measure would ...


1

For (a), use the fact that for any complex number $z$, we have the inequality $\max\{|\text{Re}z|,|\text{Im}z|\} \leq |z| \leq |\text{Re}z| + |\text{Im}z|$. For the converse direction of (b), write $F = I - D$ where $I$ and $D$ are bounded and increasing. Then if $x_0 < x_1 < \ldots x_n$, we have $$\begin{aligned} \sum_{k=1}^{n}|F(x_k)-F(x_{k-1})|&...


1

Theorem 3.27 a.) If $F\in BV$ if and only if $Re F \in BV$ and $Im F \in BV$. b.) If $F:\mathbb{R}\rightarrow \mathbb{R}$, then $F\in BV$ if and only if $F$ is the difference of two bounded increasing functions; for $F\in BV$ these functions may be taken to be $(\frac{1}{2}(T_F + F)$ and $\frac{1}{2}(T_F - F)$. c.) If $F\in BV$, then $F(x^+) = ...


1

For the first part of your question, define $$ h_n=|f|+|f_n|-|f-f_n| $$ Then $h_n\geq 0$ for all $n$, so we can apply Fatou's lemma to obtain $$2\int |f|=\int \liminf_{n\to\infty}h_n\leq \liminf_{n\to\infty}\int h_n=2\int|f|-\limsup_{n\to\infty}\int|f-f_n|$$ and since $\int|f|$ is finite this shows that $$\limsup_{n\to\infty}\int |f-f_n|=0 $$ For the ...


1

If $F\subset\Bbb Z$ is finite then $A^F$ is a finite set, so there is a uniform probability measure $\mu_F$ on $A^F$; that's the measure that assigns the same measure to each point. And there is a natural projection $\pi_F:A^{\Bbb Z}\to A^F$. The uniform measure $\mu_u$ is defined by $$\mu_u(\pi_F^{-1}(S))=\mu_F(S)\quad(S\subset A^F).$$ Asking whether ...


1

Suppose that $\{r_j\}_j$ is a sequence of positive real numbers, and $\{x_j\}_j$ is a sequence in $\mathbb{R}^n$. Suppose also that there are $r \geq 0$ and $x \in \mathbb{R}^n$, such that $$\lim_{j\rightarrow \infty}(r_j,x_j) = (r,x)$$ then is it true that $$\lim_{j\rightarrow \infty}\chi_{B(r_j,x_j)} = \chi_{B(r,x)} \textrm{ pointwise?} $$ where $...


1

As $\int |f_n - f| \to 0$, we have $\int_E |f_n - f| \to 0$ for any measurable $E$. Now: $$\left| \left| \int_E f_n \right| - \left| \int_E f \right| \right| \le \int_E| f_n - f| $$ So the LHS converges to $0$, i.e. $\int_E f_n \to \int_E f$. Now we also have: $$\int ||f_n| - |f|| \le \int |f_n - f|$$ So the LHS converges to $0$, hence by above, $\int |...


1

Yes, we can insure the desired nestedness. First: If $A,B$ are closed disjoint subsets of $ \mathbb R,$ then there exists a continuous $\varphi $ on $\mathbb R$ such that $\varphi = 1$ on $A,$ $\varphi = 0$ on $B.$ So in the problem at hand, there is a closed $F_1\subset E$ with $m(E\setminus F_1) < 1$ and a continuous $g_1$ such that $g_1=f$ on $F_1.$ ...


1

$|f_n + g_n - (f+g)| \le |f_n - f| + |g_n - g|$, so wherever $|f_n - f| < \epsilon/2$ and $|g_n-g| < \epsilon/2$ we'll have $|f_n + g_n - (f+g)| < \epsilon$. Thus $$ \eqalign{\{x: |f_n(x) + g_n(x) &- (f(x) + g(x))| \ge \epsilon\}\cr& \subseteq \{x: |f_n(x) - f(x)| \ge \epsilon/2\} \cup \{x: |g_n(x) - g(x)| \ge \epsilon/2\}}$$ so if each ...



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