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## Hot answers tagged measure-theory

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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