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6

For any $r<1$, you can construct a Cantor set with Hausdorff dimension $r$ by varying the lengths of the intervals in the usual Cantor set construction. In particular, you can let $C_n\subset[0,1]$ be a Cantor set of Hausdorff dimension $1-1/n$ for each $n$. The union $C=\bigcup C_n$ then has Lebesgue measure $0$ because each $C_n$ does, but Hausdorff ...


5

No, such a $\sigma$-algebra does not exist. Suppose it does exist; then, since $f(x)=x$ is continuous, then, for any $a\in\mathbb R$, $(a,\infty)\in\mathcal{B}$, therefore $$(a,\infty)=f^{-1}(a,\infty)\in \mathcal{F}.$$ This shows that $\mathcal{B}\subseteq\mathcal{F}$, hence $\chi_{(0,\infty)}$ is measurable, but is is not continuous. This is a ...


4

The integral diverges. To see this, we can write $$\int_0^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx=\int_0^{n/3} \left(1-\frac{3x}n\right)^ne^{x/2}\,dx+\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx \tag 1$$ We will present two parts. In Part $1$, we will show that the first integral on the right-hand side of $(1)$ converges. In Part $2$, we will ...


4

Let \begin{align}\left( \int_{-\infty}^{\infty} \sqrt{p}\sqrt{q} \;d\mu \right)^2 &= \left( \int_{-\infty}^{\infty} \min(\sqrt{p},\sqrt{q})\cdot \max(\sqrt{p},\sqrt{q}) \;d\mu \right)^2 \\ &\leq \int_{-\infty}^{\infty} \min(\sqrt{p},\sqrt{q})^2 \;d\mu \cdot \int_{-\infty}^{\infty} \max(\sqrt{p},\sqrt{q})^2 \;d\mu \\ &\leq ...


4

Well, if $f_k$ could be negative, then its integral might not even be defined. For instance, if $X=\mathbb{R}$ with Lebesgue measure and $f_k(x)=x$ for some $k$, there is no good way to define $\int f_k$ (it should morally be "$\infty-\infty$"). On the other hand, the integral of a nonnegative measurable function can always be defined (though it might be ...


4

Applying Cauchy's Mean Value Theorem twice says that there are $0\lt h_1,h_2\lt h$ so that $$ \begin{align} \left|\frac1h\left(\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right)\right| &=\left|\frac{\cos(h)-1}{h^2}\cos(x)+\frac{h-\sin(h)}{h^2}\sin(x)\right|\\ &=\left|-\frac{\cos(h_1)}2\,\cos(x)+\frac{\sin(h_2)}2\sin(x)\right|\\[4pt] &\le1 \end{align} $$ ...


4

Yes. Suppose not, look at $X^c$ which is of positive measure. By Lebesgue Density Theorem, there exists $\sigma, \tau\in 2^{\omega}$ (WLOG might assume they have the same length) such that $X$ and $X^c$ has measure $>\frac{1}{2}$ above $\tau, \sigma$ respectively. By hypothesis, $X$ has measure $>\frac{1}{2}$ above $\sigma$ too. But then above ...


3

We have to assume $\mathbb{E}(|X|^r)<\infty$; otherwise the expession $\|X_n-X\|_{L^r}$ might not even be finite. Note that $$\|X_n-X\|_{L^r}^r = \int_{|X_n-X| \leq \epsilon} |X_n-X|^r \, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X|^r \, d\mathbb{P} \tag{1}$$ for any $\epsilon>0$ and $n \in \mathbb{N}$. Obviously, $$\int_{|X_n-X| \leq ...


3

Hint: Show that $$\mathcal{D} := \{B \in \mathcal{B}(\mathbb{R}); X^{-1}(B) \in \mathcal{F}\}$$ is a Dynkin system. Conclude from the fact that $$\mathcal{G} := \{(a,b]; a<b\}$$ is contained in $\mathcal{D}$ and that $\mathcal{G}$ is a $\cap$-stable generator of $\mathcal{B}(\mathbb{R})$ that $$\mathcal{D} = \sigma(\mathcal{G}) = ...


3

The outer measure of $E$ is defined of as the infimum of the following set $$\left \{\sum_{k=1}^{\infty} \mu(E_k) \colon \{E_k\}_{k=1}^\infty \text{with $E_k\in S$ such that $E \subset \bigcup_{k=1}^\infty E_k$ } \right \}$$ Now nothing guarantees that for some set $E$ there is even one $\{E_k\}_{k=1}^\infty$ with $E_k\in S$ such that $E \subset ...


3

It isn't true. The standard counterexample is to look at $L^2((0,1))$ with Lebesgue measure and take $f_n(x) = \sqrt{2} \sin(n \pi x)$. The functions $f_n$ are orthonormal in $L^2$, so by Bessel's inequality they converge weakly to 0. But pointwise, the sequence $\{f_n(x)\}$ diverges for every $x \in (0,1)$.


3

Let $f_n(x)=\left(1-\frac{3x}{n}\right)^ne^{x/2}$. Then $$\lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)=\lim_{n\to\infty }f_n(x)\cdot \underbrace{\lim_{n\to\infty }\lambda_{[0,n]}(x)}_{=\lambda_{[0,\infty [}(x)}=\lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x).$$ Therefore $$\int \lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)\mathrm d x=\int \lambda_{[0,\infty ...


3

Alternative Solution: Take any $\epsilon>0$. We will show that $\mu\{x: \limsup_n (f_n(x))^{\frac{1}{n}}>1+\epsilon\}=0$. Define $A_n=\{x: f_n(x)> (1+\epsilon)^n\}$, and $A=\{x: \limsup_n (f_n(x))^{\frac{1}{n}}>1+\epsilon\}$. Then, $A=\limsup_n A_n$. Now, ...


3

No need for induction. If your sets are $A_1,\ldots,A_n$, then consider all the intersections $B_1\cap \cdots \cap B_n$, where each $B_i$ is either $A_i$ or $A_i^c$. There are at most $2^n$ such intersections (you may have repetitions). Added: the $\sigma$-algebra is then formed by all unions of these sets, namely all $C_1\cup\cdots\cup C_{2^n}$, where each ...


3

So $|\sigma(F)| \le 2^{2^{|F|}}$ for a finite family of sets $F$, where $|A|$ is used to denote the number of elements in $A$.


3

The difference $\gamma = \alpha - \beta$ is a signed measure. The corresponding bounded linear functional $\phi$ on $C[0,1]$ must satisfy $\phi(1) = 2$, $\phi(x) = 1$, $\phi(x^2) = 0$. Since $1$, $x$ and $x^2$ are linearly independent, this defines an affine subset of $M[0,1]$ of codimension $3$. A typical $3$-dimensional subspace of $M[0,1]$ will ...


2

In fact, $F$ is uniformly continuous in $\mathbb R$. To see this, let $\varepsilon>0$ and set $g_n(x)=\min\{\lvert\, f(x)\rvert,n\}$. By virtue of Lebesgue Dominated Convergence Theorem, $ \|g_n-\lvert\, f\rvert\|_{L^1}\to 0$. Let $N>0$, such that $\|g_N-\lvert\, f\rvert\|_{L^1}<\varepsilon/2$. Set ...


2

If $A=\bigcup_{a\in D}F_a$, then $A^c=\bigcup_{a\in \{0,1\}^n\setminus D}F_a$.


2

For a "naturally occurring" example, let $b_1$ and $b_2$ be positive integers $\geq 2$ such that no positive integer power of $b_1$ equals a positive integer power of $b_2$ (i.e. $(b_1)^m = (b_2)^n$ has no solution where $m$ and $n$ are positive integers). Kenji Nagasaka proved in 1979 that the set of real numbers normal to base $b_1$ but not normal to base ...


2

(i) if $\phi_1(x)=\phi(x)$ except on a set of measure zero $A$, then $g(x,\phi(x))=g(x,\phi_1(x))$ except on that same set of measure zero. Thus two functions which are members of the same a.e. equivalence class in $L^r$ give two functions equivalent in $L^s$. (ii) holds because $x\mapsto g(x,\phi(x))$ is a real function of $x$ defined for $x\in \Omega$ ...


2

Fix $\varepsilon$. First us tightness to find a compact subset $K=K(\varepsilon)$ of $C[0,+\infty)$ such that $\mathbb P_n(K)\gt 1-\varepsilon$. Use the uniform convergence of $(f_n)_{n\geqslant 1}$ to $f$ in order to handle the integral of $f_n$ over $K$. Use the uniform bound to handle the integral of $f_n$ over the complement of $K$ (which has a measure ...


2

You would like to prove that, $\forall E\in \mathcal{A}$, $$\mu(E\setminus \limsup_{n\to\infty}T^n(E))=\mu\left (\bigcup_{k=1}^\infty \left(E\setminus\bigcup_{n=k}^\infty T^n(E)\right)\right)=0$$ Let us prove by contradiction. Suppose $$\mu\left (\bigcup_{k=1}^\infty \left(E\setminus\bigcup_{n=k}^\infty T^n(E)\right)\right)>0$$ Then there is ...


2

Consider a Vitali set $A \subset [0,1]$, which is non measurable. The aim is to find measurable functions $f_a$ for each $a \in A$, such that $\sup\limits_{a \in A} f_a = \mathbb 1_A$. Hint :


2

Hint: Show that $$ \int_0^1 \frac{\sin x}{x^{3/2}}dx \quad\text{exists } $$ And $$ \lim_{n \rightarrow \infty} n \int_{\frac{1}{n}}^{1} \frac{\cos(x+\frac{1}{n})-\cos(x)}{x^{\frac{3}{2}}}dx=-\int_0^1 \frac{\sin x}{x^{3/2}}dx $$


2

As the Borel-$\sigma$-algebra is not complete, there is a non-measurable set $A\subseteq \mathbf R$ of outer measure zero. Let $f = \chi_A$ (the characteristic function of $A$) and $f_n = 0$. Then $f_n \to f$ almost everywhere (namely outside of $A$), then $f_n$ are Borel-measurable, but $f$ is not.


1

Sure, it is true, and you have the right proof. A minor correction is that it would be better to write $f(x)=g(x)$ instead of $f(x)-g(x)=0$. The truth is that your proof works even if the limits are $+\infty$ or $-\infty$.


1

The measure theoretic part was answered, so let me complement it by answering the choice related question. The axiom of countable choice is needed on a far more fundamental level when you talk about measure theory. It is consistent that the real numbers are a countable union of countable sets. In that case there is no $\sigma$-additive Borel measure ...


1

Note that if $f=\chi_I$ almost everywhere and $f$ is continuous, you must have for example $$f\lvert_{(-\infty,0)}=0$$ since if for some $x \in (-\infty,0)$ you have $f(x)\neq0$, then from continuity you must have that $f(x)\neq0$ on some ball $B_\epsilon(x)$. But then $f$ differs from $\chi_I$ on a set that is not a measure zero set (namely $B_\epsilon(x) ...


1

Note that it is sufficient to consider sets $A$ of the form $C\times D$, where $C$ and $D$ are open intervals (because these generate the $\sigma$-algebra). For these sets you only have to look at $f^{-1}C\cap g^{-1}D$.


1

This is my try. Let: $f_n (y) = \frac{e^y}{n^2y^4+1} \mathbb I_{[0,1]}$ , where $\mathbb I$ is indicator function. We have: $f_n(y) \to 0$, as $n \to \infty$. For integrability and domination condition, for every $n \in \mathbb N$, $|f_n(y)| \leq \frac{e^y}{y^4+1} \mathbb I_{[0,1]} \leq e^y \mathbb I_{[0,1]}$. This function is integrable on $\mathbb R$, ...



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