# Tag Info

6

What this example tells you is that your intuition about these things is not really reliable. Don't worry too much about that; everybody goes through it. In particular, the union of the intervals does not cover the entire real line. There are gaps -- they are small (none of them contain an interval), but there are a lot of them, and somehow they manage to ...

6

A general strategy in scenarios where you are trying to extend a proposition like this from a finite measure space to a $\sigma$-finite one is to break the $\sigma$-finite space in question into countably many finite-measure disjoint measurable pieces and apply the theorem to each chunk individually, and then try to stitch things back up again. To do this ...

3

Notice that your even is a zero-one event. Define the event $A_n:=\{X_n<c\}$ for some $c>0$. Then $$\sum_{n\geq 0} P(A_n)=\sum_{n\geq 0} c/n=\infty.$$ So by the reverse Borel Cantelli lemma, $A_n$ occurs infinitely often with probability 1. This implies that $P(\omega: X_n(\omega)\rightarrow\infty)<1$, and since it's a zero-one event, its ...

3

It is not too hard to show that $\overline{f_n}(x) = \sup_{k \ge n} f_k(x)$ and $\underline{f_n}(x) = \inf_{k \ge n} f_k(x)$ are measurable functions. Note that these are non increasing, non decreasing sequences respectively, hence they have measurable limits $\overline{f}(x) = \lim_n \overline{f_n}(x)$, $\underline{f}(x) = \lim_n \underline{f_n}(x)$. ...

3

Let $U$ be a random variable which is uniform on $(0,1]$. Let $X=Y=U^{-1/2}$. Then $E[X]=E[Y]=2$ but $E[XY]=+\infty$. You can make lots of examples of functions like this, which are not integrable but their square root is integrable, because it diverges "more slowly". On an infinite measure space you can also have the opposite phenomenon with tailing: ...

3

Because $t$ is fixed and will play no role, I'll drop it from the story. Thus we have an integrable random variable $X$ such that $E[\varphi(X)|\mathcal F]=\varphi(E[X|\mathcal F])$ for all bounded and continuous $\varphi$. I will allow $\varphi$ to be complex-valued. (Look at the real and imaginary parts of $\varphi$ separately and then add.) Let's define ...

3

You are allready there. Observe that: $$\int |f|dm\geq\int |f|1_Adm=(+\infty)\times m(A)=+\infty$$ Showing that $f$ is not integrable. This is proved on base of the assumption that $f$ is not finite a.e..

2

I have also thought about the question concerning the associativity of the $\sigma$-algebras and will present another proof, which does not make use of the $\pi - \Lambda ~ -$ system Theorem but uses the direct image $\sigma -$algebra: As mentioned let $(X_i, M_i, \mu_i)$ for $i = 1, 2, 3$ be $\sigma-$finite measure spaces. Then using the same ...

2

Hint: $$\left|\int f_n - \int f\right| \leq \int |f_n - f|$$ and the quantity on the right can be controlled because of uniform convergence and the fact that $\mu(X) < \infty$. To prove $f\in L^1$, use something similar such as $$\int |f| \leq \int |f - f_N| + \int |f_N|$$ for some fixed $N \in \mathbb{N}$ large enough.

2

Let $f \in L^{p_1}(E)$ bounded. Then there exists $M\geq 0$ such that $\sup_{x \in E} |f(x)| \leq M$ and $\|f\|_{p_1}^{p_1}=\int_E |f(x)|^{p_1}\,dx < \infty$. Case 1: $p_2<\infty$ For $p_2 > p_1$, we have \begin{align} \|f\|_{p_2}^{p_2}& =\int_E |f(x)|^{p_2}\,dx \\ &= \int_E |f(x)|^{p_2-p_1}|f(x)|^{p_1}\,dx \\ & \leq \left(\sup_{x ...

2

If $\Bbb E[1_A|\mathscr F_t]=1$, then $\Bbb E[1_A]=1$, which is the same as $1_A=1$. In this case $\Bbb E[1_A|\mathscr F_s]=1$ (almost surely) for each $s$. Likewise, if $\Bbb E[1_A|\mathscr F_t]=0$, then $\Bbb E[1_A]=0$, which is the same as $1_A=0$. In this case $\Bbb E[1_A|\mathscr F_s]=0$ (almost surely) for each $s$. If $\Bbb E[1_A|\mathscr F_t]=p$, ...

2

The purpose was to define integrals in such a way that it is easy to prove that in some circumstances $\displaystyle\lim\limits_{n\to\infty} \int_A f_n(x)\,dx = \int_A \lim\limits_{n\to\infty} f_n(x)\,dx$ and things like that. If there are some other purposes that it doesn't serve, that's not a criticism of it. You cannot abandon the Archimedean property ...

2

Yeah, your proof is correct. Regarding your first question concerning integrability: If you prefer, you can restate the result as follows. The following statements are equivalent: $\mathcal{F}$ and $\mathcal{G}$ are independent. For all bounded $X_F \in m \mathcal{F}$ and bounded $X_G \in m \mathcal{G}$, it holds that $\mathbb{E}(X_G X_F ) = ... 2 I think that you must do some changes before using DCT. a) Change of variable$u=x^n$. Your integral is now $$I_n=\int_0^{+\infty}\frac{\sin(x^n)}{x^2}dx=\frac{1}{n}\int_0^{+\infty}\frac{\sin(u)}{u^{1+1/n}} du$$ b) Integration by parts, using$(1-\cos(u))^{\prime}=\sin(u)$: $$I_n=\frac{n+1}{n^2}\int_0^{+\infty}\frac{1-\cos(u))}{u^{2+1/n}}du$$ And now ... 2 Let$q\in\mathbb{Q}$and$n\in\mathbb{N}$be arbitrary. Then, for each$k\in\mathbb{N}$, the event$A_{q,n}^{\left(k\right)}:=\left\{ \left|X_{k}-q\right|<\frac{1}{n}\right\} $has the probability$\mathbb{P}\left(A_{q,n}^{\left(k\right)}\right)=\mathbb{P}\left(A_{q,n}^{\left(1\right)}\right)>0$which is independent of$k\in\mathbb{N}$. Since the ... 2 For any$a>0,\lim_{x\to 0^+}x^a\ln (1/x) =0.$So given$0<p<\infty,$we have for small positive$x$$$x^{1/(2p)}\ln(1/x) < 1 \implies x^{1/2}(\ln(1/x))^p < 1 \implies (\ln(1/x))^p < x^{-1/2}.$$ Since$\int_0^1 x^{-1/2}\, dx < \infty,$we have the result. 2 You can do this change of variables : u =$\log{1/x} \implies x = e^{-u}$So, to calculate the p-norm, consider the following integral$\int_0^{1}\log(\frac{1}{x})^pdx = \int_{0}^{\infty} u^p e^{-u} du$The integral in the r.h.s is the well-known gamma function$\Gamma(p+1)$, which converges for the values of p that we are interested. For the ... 2 Later hint,$p=2$: Let$\epsilon >0.$For$0<x<\epsilon$we have $$|f(\epsilon)-f(x)| \le \int_x^\epsilon |f'(t)|\, dt \le (\int_x^\epsilon|f'(t)|^2t\, dt)^{1/2}(\int_x^\epsilon t^{-1}\, dt)^{1/2}$$ $$\le (\int_0^\epsilon|f'(t)|^2t\, dt)^{1/2}(\int_x^\epsilon t^{-1}\, dt)^{1/2}.$$ Note the first integral on the last line$\to 0$as$\epsilon \to ...

2

It is a quotient space! $\mathbb{R}$ here should be thought of as the one-dimensional subspace of $L^2(\Omega)$ which consists of the (a.e.) constant functions. The norm given is the canonical norm on the quotient of a Banach space mod a closed subspace. This construction is discussed, for instance, in section III.4 of Conway's A Course in Functional ...

2

The only point where the measures itself intervenes is in the definition of strong measurability, through simple convergence $\mu$-almost everywhere. A $\sigma$-finite measure is equivalent (in the sense has the same negligible sets) to a finite measure: partition $\Omega$ into countably many sets of finite $\mu$-measures, and define $d\nu=\rho\ d\mu$ where ...

2

I get your confusion, but a set having "gaps" does not imply the set misses an open interval. For more trivial examples, consider $\mathbb R\setminus\{\sqrt{2}\}$ or $\mathbb R\setminus\mathbb Q$. The first set has a single gap but still contains every rational number. The second set has gaps that are dense in the whole real line, but have zero measure. The ...

2

The argument consists of three steps, as outlined by Frank Science. Restrict attention to the ball $B_n=\{x:\|x\|\le n\}$, and observe that $f$ is Lipschitz continuous on $B$ (since its gradient is bounded there). Apply the result that Lipschitz maps preserve sets of measure zero: see Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map ...

2

For fixed $k \in \mathbb{N}$ we have $$e^y = \sum_{n \geq 0} \frac{y^n}{n!} \geq \frac{y^k}{k!} \qquad \text{for all y>0}.$$ In particular, if $y = x^{\alpha}$ for some $x > 0$, then $$e^{x^{\alpha}}\geq \frac{x^{k \alpha }}{k!},$$ i.e. $$e^{-x^{\alpha}} \leq \frac{k!}{x^{k \alpha}}.$$ Choose $k= k(\alpha)$ sufficiently large such that ...

2

The Fourier transform is a linear map, you only have to check that $$\forall f\in L^1(\mathbb R), \mathcal F(f)=0 \Rightarrow f=0.$$ Let $f\in L^1(\mathbb R)$ such that $\mathcal F (f)=0$. Hence $\mathcal F(f)$ is $L^1(\mathbb R)$ since its the zero function and therefore its Fourier inverse exists. So $$f=\mathcal F^{-1}(\mathcal F(f))=\mathcal ... 1 As stated in your question, the "corollary" is useless: you did not say anything about the covering of A by B_i. The existence of a ball of measure equal to \mu(A), somewhere in the space, is of little consequence. An important point is that such balls cover A up to a subset of measure zero. Let's apply the covering lemma to the measure \mu+ ... 1 A measure is defined only on measurable sets, so your first question doesn't really makes sense. About the exemple, the most common space in integration, the Borel sigma algebra of \Bbb R with the Lebesgue measure is not complete. The completion of this measure is the space of the Lebesgue sigma algebra 1 It can be shown that \mathbb{R} endowed with the Borel \sigma-algebra is not complete. 1 Let \Omega be a set with at least two elements and \mathcal{S} = \{\emptyset, \Omega\}; (\Omega, \mathcal{S}) is a measurable space. Now let \mu be the zero measure on (\Omega, \mathcal{S}). The space (\Omega, \mathcal{S}, \mu) is not a complete measure space. No proper nonvoid subset of \Omega is measurable. For any measure space ... 1 You have shown that \rho(A,B) \geq 0 because P is a probability measure and thus P(A\Delta B)\geq 0. Likewise since the symmetric difference is the empty set if and only if the sets are equal, therefore \rho(A,B)=0 \iff A=B. You have shown that symmetry of \rho follows from the symmetry of symmetric difference; via \rho(A,B) = P(A\Delta B) = ... 1 No, there is nothing to be said about the Cesàro averages of the whole sequence. They can be as bad as the sequence itself. Indeed, given any weakly convergent sequence \{f_n\}, we can consider another sequence \{g_n\} defined as$$f_1,f_1, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_3, \dots  where the term $f_n$ appears ...

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