# Tag Info

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First, assume that $f$ is an indicator function, say $f=\mathbf 1_A$ where $A$ is a measurable. You can find Borel sets $B,C$ such that $B\subseteq A\subseteq C$ and $C\setminus B$ has measure $0$; then take $g:= \mathbf 1_B$ and $h:=\mathbf 1_C$. Next, assume that $f$ is a simple function, say $f=\sum_{i=1}^N \alpha_i\mathbf 1_{A_i}$, where the sets $A_i$ ...

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As mentioned by Svetoslav, the key word here is "uniform integrability". First observe that the result holds true if you take just one function $f$ instead of a sequence (or if you prefer, when your sequence $(f_n)$ is constant). This is not completely trivial if you do not want to use any theorem. For every $k\in\mathbb N$, set $E_k:=\{ 1/k\leq \vert ... 1 Let $$A:=\left\{X_1=x_1,\ldots,X_n=x_n\right\}\;.$$ As Calvin Khor pointed out, we can assume, that$x_i\ne x$. Thus, $$\left\{\tau_x^1<\infty\right\}\cap A=\underbrace{\left\{\exists k>n:X_k=x\right\}}_{=:B}\cap A\;.$$ Let $$\tilde X:=\left(X_{k+n}\right)_{k\in\mathbb N_0}$$ and$f$be the indicator function of$\bigcup_{n\in\mathbb ...

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I think we cannot prove integrability without further assumptions on the $X_n$s and $Y_n$s. If $\varphi$ is a concave function and $X,Y\in L^1$, then $\varphi(X),\varphi(Y)\in L^1$ by Jensen's inequality. However, the exponential function is convex. For instance, if the PDF of $X$ is $f_X(x)=e^{-x}$, supported on $\mathbb{R}^+$, the PDF of $Z=e^X$ is ...

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I believe this is what they had in mind. For brevity $A_n:=[X_1=x_1,…,X_n = x_n]$. \begin{align}P_x(τ_x^1 = ∞ ∩ A) &= P_x(τ_x^1 = ∞ |A) P(A) \\&\overset{\star}{=} P_x(τ_x^1 = ∞ |X_n = y)P(A) \\&= P_y(τ_x^1 = ∞)P(A) \end{align} Where $\star$ is where the markov property is needed. I'm not sure how to fully justify this. Regarding the choice of ...

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I'm not sure why you need that step. You want $l(I) \le m^*(I).$ You can get that by showing that if $I\subset \cup I_n,$ then $l(I) \le \sum l(I_n).$ Here $\{I_n\}$ is an open cover of $I.$ Hint: $I$ is compact, so the open cover $\{I_n\}$ of $I$ has a finite sub cover.

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go through the definition of outer measure. that'll help you out. it's infimum over the covers that covers $I$. now as that is infimum hence there'll be a cover such that $\sum_{n=1}^{\infty}l(I_n) \leqslant m^*(I)+\frac{\epsilon}{2}$

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Any continuous function on a compact set is automatically bounded. Since $L^2[0,T]$ is the completion of the continuous functions on $[0,T]$ with the norm $\|f\|_2 = \sqrt{\int_0^T |f(t)|^2 dt}$, it follows that the continuous functions (or equivalence classes thereof) are dense in $L^2[0,T]$. Alternative: Choose $f \in L^2[0,T]$. Let $f_M(t) = ... 2 Say that$|f(x)| \le M$is not satisfied a.e. and pick$\epsilon > 0$and$\delta > 0$such that if$G = \{|f(x)| \ge M + \epsilon\}$then$\mu(G) \ge \delta$. Now notice that $$0 = \lim_{n \to \infty}\mu(\{|f(x) - f_n(x)| \ge \epsilon\}) \ge \lim_{n \to \infty}\mu(\{|f(x)| - |f_n(x)| \ge \epsilon\}) \ge \mu(G) \ge \delta.$$ You can apply the ... 0 I've noticed that there are a bunch of questions from the Calculus of Variations course that I'm taking. Please stop this. The policy doesn't say that you're allowed to ask questions here. If you're stuck on the problems, talk to the professor and your fellow classmates. 1 A function$f: (X,\mathcal{A}) \to (Y,\mathcal{B})$is measurable if$f^{-1}(B) \in \mathcal{A}$for all$B \in \mathcal{B}$. Since the function$x \mapsto ax$is continuous, what can you say about$f^{-1}(B)$? 1 The proof is quite similar to the proof of part a). We consider a sequence$x_n \to x_0$such that$x_n \neq x_0$for all$n > 0$, and the difference quotient $$\frac{F(x_n) - F(x_0)}{x_n - x_0} = \int_Y \underbrace{\frac{f(x_n,y) - f(x_0,y)}{x_n - x_0}}_{g_n(y)}\,d\mu(y).$$ By the mean value theorem, for every$n$, and every$y$, there is a$\xi_n(y)$... 0 One easy counterexample is something like$\mathbb{R}^2$with Lebesgue measure (or any measure you like that is translation/rotation-invariant and measures intervals in the usual way): it's$\sigma$-finite since you can cover it with the disks$\{ x:\lvert x \rvert < n \}$, but the sets$[n,n+1) \times \mathbb{R}$form a disjoint cover of infinite-measure ... 1 It says "there is" only of course. Take your$E$,$\sigma$-finite but not with finite measure, and put $$E_i = \begin{cases} E & \text{if i=0} \\ \emptyset & \text{otherwise}\end{cases}.$$ 2 It could happen that some decomposition of a$\sigma$-finite set into countably many disjoint subsets contains infinite measure sets. For instance, work in$\mathbb{R}$with$E_1=(-\infty,0)$and$E_k=[k-2,k-1)$for$k \geq 2$. 0 The definition is just "there exists". The other way is totally not true. Consider$\Bbb{R}$with the usual measure. Then: $$\bigcup\limits_{i=1}^{\infty} \Bbb{R} = \Bbb{R}$$ but$\mu (\Bbb{R})=\infty$Edit: I just saw that you say "disjoint". Well this is still not true. Consider$E_n=\Bbb{R}$for$n=1$and$E_n=\emptyset$otherwise. There are tons of ... 2 You are on the right track. Look at the extreme cases for (A) and (B). (A) If$C=\varnothing$, then$P(A\cup B)=0.7$. If$A=B=\varnothing,P(C)=0.7$, then$P(A\cup B)=0$. So $$0\le P(A\cup B)\le0.7$$ (B) If$C=A\cup B$, then$(A \cup B) \cap C^c=\varnothing$so$P((A \cup B) \cap C^c)=0$. If$C=\varnothing$, then$P(A\cup B)=0.7$and$(A \cup B) \cap ...

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Since Hilbert spaces are first-countable, it will be enough to check continuity using sequences (as opposed to using nets). We shall base our proof on Lebesgue's dominated convergence theorem (for which using sequences and not nets is essential, as shown here and here). Let $x_n \to x$ in $H$. Note that the function y \mapsto \Bbb e ^{\Bbb i \langle x_n, y ... 1 It should be \begin{align} \int_{\mathbb R^2} f(x,y)&=\int_{\mathbb R}\left(\int_{\mathbb R}\chi_{(A-x) \cap B}(y)dy\right)dx \\ &=\int_{\mathbb R}\left(\int_{B}\chi_{(A-x) }(y)dy\right)dx \\ &=\int_{\mathbb R}\left(\int_{B}\chi_{A }(y+x)dy\right)dx \\ &=\int_{B}\left(\int_{\mathbb R}\chi_{A }(y+x)dx\right)dy \\ &=\int_{B}m(A)dy \\ ... 0 LetE=\{x: f(x) = \infty\}$and$E_n=\{x : f(x) > n\}$for$n\in\mathbb N$. Then$\mu(E_1)<\infty$(as otherwise$\int f\ \mathsf d\mu =\infty$), and$E_n\supset E_{n+1}$, so by continuity from above, $$\mu(E) = \mu\left(\bigcup_{n=1}^\infty E_n\right) = \lim_{n\to\infty} \mu(E_n).$$ For each$n$, we have $$\int f\ \mathsf d\mu \geqslant \int_{E_n} ... 4 Say \alpha is the supremum of \mu(F) over all F\subset E with \mu(F)<\infty. If \alpha=\infty you're done. Suppose \alpha<\infty. Then \alpha-1/n<\alpha, so for every n there exists F_n\subset E with \alpha-1/n<\mu(F_n)<\infty. Let S_n=\bigcup_{j=1}^n F_j. Then \mu(S_n)>\alpha-1/n. Also S_n\subset E and ... 0 Here's an example. S is the unit square [0,1]\times[0,1], \Sigma is the Borel \sigma-algebra on S, and \mu is Lebesgue measure (area) on (S,\Sigma). \tilde\Sigma is the \sigma-algabra of subsets of S of the form B\times[0,1], where B is a Borel subset of [0,1]. Lets take f(x,y)=x^2 and h(x,y)=x for all (x,y)\in S. Both f and ... 2 I think I have a counterexample for general outer measures. Let X be a set with at least 3 elements x,y,z. One can just use X = \{x,y,z\}. Define a finite outer measure \mu^* : \mathcal{P}(X) \to [0,\infty] by$$ \mu^*(E) = \begin{cases} 0 & \text{ if } E = \varnothing \\ 1 & \text{ if } |E| =1 \text{ or } 2 \\ 2 & \text{ if } |E| \geq ... 2 Pushing all the way to the extreme, just a single function can do all the job: Let$g_i(x)$by $$g_i(x) = \frac{1}{2\sqrt{x}} \mathbf{1}_{(0, 4^{-i})}(x).$$ Notice that$g_i$is unbounded on any neighborhood of$0$and its integral over$\Bbb{R}$is$2^{-i}$. Now enumerate$\Bbb{Q}$as$\{r_1, r_2, \cdots\}$and define$f$by $$f(x) = ... 1 Let q_1,q_2, \dots be the rationals. Define f(x) = \chi_{(0,1)} + \sum_{k=1}^{\infty}n\chi_{\{q_n\}}. The sequence f,f,f,\dots then does the job. 2 I don't know the answer in the case of a general outer measure. But most reasonable outer measures arise as in the Caratheodory construction (cf. https://en.wikipedia.org/wiki/Pre-measure and https://en.wikipedia.org/wiki/Carath\%C3\%A9odory\%27s_criterion), i.e. we have a (semi)ring R of subsets of X and a premeasure \mu_{0}:R\to\left[0,\infty\right] ... 2 Let h(x) = \frac{1}{\sqrt{\pi}}\exp{(-x^2)}. I pick this because \int_{-\infty}^{\infty}h(x)dx = 1. Let g_n(x) = \begin{cases} 0 &\mbox{if } x \in \mathbb{R} \backslash \mathbb{Q} \\ n & \mbox{if } x \in \mathbb{Q} \end{cases} Let f_n := g_n + h This works as \mathbb{Q} is dense in \mathbb{R} and has Lebesgue measure 0, so f_n is ... 1 Here's an example to show that your concern about isolated points is justified. Let f(x)=0 for all x\in{\Bbb R} and g(x) = \max(x,0). These two functions are continuous and the are equal at (Lebesgue) a.e. point of B:=(-\infty,0]\cup\{1\}. But they are not identically equal on B. 7 Let f_0 be function unbounded in a neighborhood of 0 with finite integral, e.g.$$f_0(x) = \frac{1}{\sqrt{|x|}} \chi_{[-1,1]}.$$You may need to rescale to make sure it has integral 1. If \{r_n\} is an enumeration of the rationals (or any dense countable set) you can define$$f_n(x) = f_0(x-r_n)$$so that f_n has integral 1 but is unbounded on ... 3 To speak of a property holding a.e. on a set X you first need a measure. If P(x) is a property defined for every point x \in X, then P is true \mu-almost everywhere if$$\mu(\{ x \in X \mid \neg P(x)\}) = 0.$$Perhaps a bit more precisely, you might require that there is a \mu-measurable set N with the property that \mu(N) = 0 and \{ x \in X ... 1 Suppose for now \mu real. Let |\mu|^{\perp} the pushforward of |\mu| with respect to the projection P^{\perp}:H\rightarrow V^{\perp}. Using the hypothesis and a change of variables, we get that for every y\in V^{\perp},$$\widehat{|\mu|^{\perp}}(y) = \int \limits_{V^{\perp}}e^{i(x,y)}d|\mu|^{\perp}(x) = \int \limits_He^{i(P^\perp x,y)}d|\mu|(x) = ... 0 No, that is not the case and there are plenty of counterexamples. It is not legal to conclude: $$f=g$$ purely on base of: $$\int fd\mu=\int gd\mu$$ 2$f$is measurable as a pointwise limit of measurable functions. For the integrability of$f$: For big enough$n\in\mathbb N$because of the uniform convergence we have$|f_n-f|\leq \epsilon\Rightarrow |f|-|f_n|\leq\epsilon$. Therefore $$\int\limits_{a}^{b}{|f|d\mu}\leq \int\limits_{a}^{b}{(|f_n|+\epsilon)d\mu}$$ and we conclude that$f$is summable. ... 2 You can use upper and lower integrals for$f$, if$f-f_n<\epsilon$you have:$\int f_n dx+\epsilon m \ge \int^* fdx \ge\int_* fdx \ge \int f_n dx-\epsilon m$where$m$is the measure of the set you're integrating over. Since$\epsilon$is arbitrarily small you have$\int^* fdx = \int_*f dx$and consequently$f$is integrable. The rest should be straight ... 1 In order to use the strong Markov property$(2)$at time$\tau:=\tau_y^{k-1}$, we need to find a bounded,$\mathcal E^{\otimes\mathbb N_0}$-measurable function$f:E^{\mathbb N_0}\to\mathbb R$with $$f\circ\tilde X=1_{\left\{\tau^+<\infty\right\}}\;,$$ where$\tilde X:=\left(X_{\tau+n}\right)_{n\in\mathbb N_0}$and$\tau^+:=\tau_y^k$. Since ... 2 You have the "pointwise" inequality $$ab \le \frac{a^p}{p} + \frac{b^q}{q}$$ whenever$a, b, \ge 0$and$\frac 1p + \frac 1q = 1$. Then $$|f| \cdot |g| \le \frac{|f|^p}{p} + \frac{|g|^q}{q}.$$ Now we use the general fact: If$F$,$G$are two integrable function so that$F\le G$and$\int F = \int G$, then$F = G$almost everywhere. Now let$F = |f|\cdot ...

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In general, it does not hold. Consider for example the sequence $$u_n = - 1_{(n,n+1)}.$$ Then $u_n \leq 0$ and $u_n \to 0 =:u$ pointwise, but $\int u_n \, dx = -1 \not\to 0=\int u \, dx$. One can certainly modify the assumptions slightly to achieve a positive result. Assume for example that $u_n \geq v$ for all $n$ and some integrable function $v$. Then ...

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For open sets, the product is also open. For the case of null sets, use the definition of the Lebesgue outer measure to show that of $E$ is null, then so is $[-N,N]^m\times E$ for arbitrary $N$. Then take countable unions.

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I read $V(\nabla u)-V(\nabla u)$ as the function defined by $V((\nabla u)(\xi))$. So if $1<p<2$, then we have from the second inequality in $$c^{-1}(\left|\xi\right|^{2}+\left|\eta\right|^{2})^{\frac{p-2}{2}}\left|\xi-\eta\right|^{2}\leq\left|V(\xi)-V(\eta)\right|^{2}\leq ... 1 Using (3) in the form$$ |V(\xi) - V(\eta)|^2 \leq K(|\xi|^2 + |\eta|^2)^{\frac{p-2}{2}} |\xi - \eta|^2 \le K (|\xi| + |\eta|)^{p-2} |\xi - \eta|^2 $$we get from (1)$$ \int_B |V(\nabla u) - V(\nabla v)|^2 \le K \int_B (|\nabla u| + |\nabla v|)^{p-2} |\nabla u- \nabla v|^2 \leq \int_B |\nabla u|^p - |\nabla v|^p$$and an application of (2) yields the ... 2 Any bijection from \mathbb{R} to itself has measurable preimages of singletons. So a straightforward way to do it is to biject a nonmeasurable set E with an interval I, and then biject \mathbb{R} \setminus E with \mathbb{R} \setminus I. Then the preimage of the interval is E which is certainly not measurable. What Borel set has nonmeasurable ... 0 Nice example. Each f^{-1}(a) is a singleton (hence measurable) but f is not measurable. [Added: As @John Ma notes, f^{-1}(a) need not be a singleton, but does have cardinality at most 2. To see the non-measurability of f, let g denote the identity function; then f=g\cdot (2\cdot 1_E -1), where 1_E is the indicator function of E. If f ... 0 Fatous's lemma in its full form has lim sup involved. What's the limit of \chi_{(-1)^n}? It doesn't exist! But the measure of these two functions is the same so the LHS of this equation does hold. 1 Take f_j:\mathbb N\to[0,\infty] be given by f_j(i)=a_{ij}. It is measurable, because you can take the discrete topology on \mathbb N. Then$$ \sum_{i=1}^\infty\sum_{j=1}^\infty a_{ij} =\sum_{i=1}^\infty\sum_{j=1}^\infty f_j(i)=\int_{\mathbb N} \sum_{j=1}^\infty f_j =\sum_{j=1}^\infty \int_{\mathbb N}f_j =\sum_{j=1}^\infty\sum_{i=1}^\infty a_{ij} $$1 Hints: By the monotonicity of the integral,$$\int u_n \, d\mu \leq \int u \, d\mu$$for each n \in \mathbb{N}. This implies$$\limsup_{n \to \infty} \int u_n \, d\mu \leq \int u \, d\mu. \tag{1}$$By assumption, u_n is non-negative. Using Fatou's lemma and$$\int u \, d\mu = \int \liminf_{n \to \infty} u_n \, d\mu$$show that$$\int u \, d\mu \leq ...

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EDIT: When I wrote this answer I mistakenly thought the question was about outer measure on $\Bbb R$, not $\Bbb R^n$. Things are not quite so trivial in $\Bbb R^n$. There will be a version for $\Bbb R^n$ appearing here when I have time. Thanks to Nate Eldrege for pointing out the problem. This is obvious. Say $m_*(E)=\alpha$. By definition $\alpha$ is the ...

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In addition to the "approximation by simple functions" approach, one can use the monotone class theorem for functions, as found for example here. The conditions of the theorem quoted there are met by taking the $\pi$-system to be your ${\mathcal A}$ and the vector space ${\mathcal H}$ to be the class of bounded ${\mathcal A}$-measurable functions ...

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There is a rather famous construction of such a set. It helps to know the following facts: if $O$ is a nonempty open set then there exists a nowhere dense closed set $C \subset O$ with the property that $0 < m(C)$, and if $C_1,\ldots,C_n$ is a collection of closed nowhere dense sets and $O$ is a nonempty open set, then $O \setminus (C_1 \cup \cdots ... 2 I think there are two things here which our intuition needs to get used to: There are just so much more irrationals than rationals in$[0..1]$. These balls used to cover the rationals in$[0..1]$in such a construction are getting really small, and yet they are all over the place. Whenever I try to visualize the construct, my brain gets trapped in the ... 1 I don't think a sane persion would easily visualize such a construct. The construct will be a dense set with just isolated points cut out (since$\mathbb Q$is dense). The construct is to simply enumerate the rational numbers and surround them with an open set such that$m(V_n) = \epsilon k^{-n}$. Since$\bigcup V_n\$ will cover all rational numbers and ...

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