Suppose that $(S,\Sigma,\mu)$ is a finite measure space. I am trying to show that a sequence of functions $(f_n)$ converges in measure to f iff   \begin{align*} \lim_{n\to \infty} \int \frac{\left\vert {f_n - f} \right\vert}{1 + \left\vert { f_n - f} \right\vert} d\mu = 0. \end{align*}  I have shown if  \begin{align*} \lim_{n\to \infty} \int \frac{\left\vert {f_n - f} \right\vert}{1 + \left\vert { f_n - f} \right\vert} d\mu = 0, \end{align*} then $f_n \to f$ in measure. However, I am not sure how to show the reverse direction. I know that since I converge in measure that there exists a subsequence $(f_{n_k})$ that converges $\mu$-a.e. on S to f. So I can apply the monotone convergence theorem to the subsequence and get the desired result. I am not sure though how to show it for the entire sequence.
15358
Let $(X,\mathcal{M},\mu)$ be a measure space. We say that $\{f_n\}$ converges to $f$ in measure if, for any $\epsilon&gt;0$   $$ \lim\limits_{n \to \infty} \mu\Big(\{x\in X: |f_n(x)-f(x)| \ge \epsilon  \}\Big)=0.$$         Show that $\{f_n \}$ converges to $f$ in measure if $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0.$$   Suppose $\mu(X) &lt; \infty$. Show that    $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0$$   if $\{ f_n \}$ converges to $f$ in measure.
57303
I'm studying measure theory, and I came across a theorem that says that, given two $\sigma$-finite measure spaces $\left(X,\mathcal M,\mu\right)$ and $\left(Y,\mathcal N,\nu\right)$, and a set $E\in \mathcal M\otimes\mathcal N$, the function $x\mapsto \nu\left(E_x\right)$ is measurable.  The definition of the product-$\sigma$-algebra $\mathcal M\otimes\mathcal N$ is the $\sigma$-algebra generated by the rectangles $A\times B$ with $A\in\mathcal M$ and $B\in\mathcal N$. Also, the $x$-section $E_x$ is defined as $E_x=\left\{y\in Y:\left(x,y\right)\in E\right\}$.  My problem is that I can't even convince myself that, given $E\in\mathcal M\otimes \mathcal N$, the sections $E_x$ are in $\mathcal N$ for all $x$. It looks like it has something to do with unitary sets being measurable (which shouldn't be necessary, I think), but I can't prove it even assuming that $\left\{x\right\}\in\mathcal M$ for all $x$.
37572
I'm preparing for a test for real analysis and I came across this problem in Royden's book:  Let $\{f_n\}$ be a sequence of real valued functions on $[a,b]$ that converges pointwisely on $[a,b]$ to the real valued function $f$. Show that $TV(f) \leq \liminf  ~TV(f_n)?$  This looks quite similar in form to Fatou's Lemma to me, but can't find any way to establish TV with integration, can anybody please help?  (TV is short for total variation)
77932
I am trying to solve following exercise from Folland,     If $\mu$ is a semifinite measure and $\mu(E) = \infty$, for any $C &gt; 0$, $\exists$ $F \subset E$ with $C &lt; \mu(F) &lt; \infty$.   It seems to follow from definition of semifinite measures, which you can find , but I couldn't prove it.
35734
Let $X=\mathbb{R}$ and let $\mathcal{S}=\{A\subseteq \mathbb{R}:$ either $A$ is countable or $\mathbb{R}\setminus A$ is countable }. Let $f:\mathbb{R}\rightarrow [0,\infty)$ be any function, and define $\mu_{F}:\mathcal{S}\rightarrow[0,\infty)$ by $\mu_{f}(A)=$sup $\{\Sigma_{x\in F}f(x):F\subseteq A$ is finite$\}$ Show that $\mu_{f}$ is a measure on $(\mathbb{R},\mathcal{S})$. Now I know that a measure on $(X,\mathcal{S})$ is a function $\mu:\mathcal{S}\rightarrow[0,\infty]$ such that $\mu(\bigcup_{k=1}^{\infty}E_{k})=\Sigma_{k=1}^{\infty}\mu(E_{k})$, but I am unsure of how to incorporate the disjoint sets into the definition for $\mu_{f}(A)$. I know that if $E_{1}\cup E_{2}\cup...=A$ where each $E_{i}$ is disjoint then we have \begin{equation*} \begin{split} \mu_{f}(E_{1}\cup E_{2}\cup...)=\mu_{f}(A)&amp;=\text{sup}\{\Sigma_{x\in F}f(x):F\subseteq A \space  \text{is finite}\}\\ &amp;=\text{sup}\{f(x_{1})+...f(x_{n}):\{x_{1},...,x_{n}\}\in F\} \end{split} \end{equation*} and somehow the last line should equal $\mu_{f}(E_{1})+\mu_{f}(E_{2})+...$ . Any help would be appreciated. Thank you
44569
I am having some conceptual difficulties with almost everywhere (a.e.) convergence  versus convergence in measure.   Let $f_{n} : X \to Y$. In my mind, a sequence of measurable functions $\{ f_{n} \}$ converges a.e. to the function $f$ if $f_{n} \to f$ everywhere on $X$ except for maybe on some set of measure zero.   However, the definition of convergence in measure is that $f_{n} \to f$ in measure if for every $\varepsilon&gt;0$  \begin{equation} \mu(\{x:|f_{n}(x)-f(x)|\geq\varepsilon\})\to0 \text{ as } n\to \infty \end{equation}  Heuristically, it seems to me that convergence in measure is just saying that the subset of $X$ for which $f_{n}$ does not converge to $f$ must have a measure of zero.   I cannot see the difference between a.e. convergence and convergence in measure. If anyone can point out my conceptual error it would be much appreciated.
80367
Let $\{\mathcal{F}_n\}_n$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$($\mathcal{F}_{n+1}\subset\mathcal{F}_n$) and let $\{X_n\}_n$ be a backward submartingale($E[|X_n|]&lt;\infty$, $X_n$ is $\mathcal{F}_n$ measurable and $E[X_n|\mathcal{F}_{n+1}]\geq X_{n+1}$).  If we know $l=\lim_{n\rightarrow\infty}E[X_n]&gt;-\infty$, how to prove that $\{X_n\}_n$ is uniformly integrable? Many thanks!  I have an idea, it is enough to show:    $\sup_nE[|X_n|]&lt;\infty$;  $\forall\epsilon&gt;0$, $\exists\delta&gt;0$ s.t. $\forall A\in\mathcal{F}$ such that $\mathbb{P}(A)&lt;\delta$, we have $\sup_nE[1_{A}|X_n|]&lt;\epsilon$.    For $|X_n|=2X_n^+-X_n$, we have  $$E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_0^+]-l\leq 2E[|X_0|]-l$$  which implies $\sup_nE[|X_n|]&lt;\infty$.  Secondly, $\forall A\in\mathcal{F}$, we can suppose without loss of generality that $A\in\mathcal{F}_0$.  $$E[1_{A}|X_n|]=2E[[1_{A}X_n^+]-E[[1_{A}X_n]$$  Does someone have an idea for this term? Thanks a lot.....
98512
I want to show that if $f$ is non-increasing and $f\in L_{1}([a,\infty),m)$ where $m$ is Lebesgue measure then $\lim_{t\to\infty} t f(t)=0$. So far I've been able to show that $f\geq 0$ and that $\lim_{t\to\infty} f(t)=0$. Since monotone functions are differentiable a.e. I thought about using integration by parts but couldn't get anywhere with that. Any hints or suggestions would be greatly appreciated.
9432
Let $\{s_n\}$ be an increasing sequence of step functions which converges pointwise on an interval $I$ to a limit function $f$.  If $I$ is unbounded and if $f(x) \geq 1$ everywhere on $I$ except on a set of measure zero, prove that the sequence $\{\int_I s_n\}$ diverges.  (Here $\int_I s_n$ refers to the Lebesgue Integral.)  Conceptually, this makes sense.  Clearly, if $f(x) \geq 1$ almost everywhere on an unbounded interval, then the sequence of Lebesgue integrals should diverge.  However, in proving this, I am running into issues with the convergence of $\{s_n\}$ being pointwise.  If it were uniform, I could easily show this.  How do I show that (even for pointwise convergence) I can choose $n$ large enough to make $\int_I s_n$ as big as I want?
102050
Let $(X,S,\mu)$ be a measure space. Suppose $\mu(X)&lt;\infty$. Put $E_n=\{$ $x\in X:$ $n-1\leq |f(x)|&lt;n \}$ Then, $f\in \mathcal{L}^p$ if and only if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$. My attempt: Notice, $X=\bigcup_{n=1}^{\infty}E_n$. Since $(E_n)_n$ are pairwise disjoint, we deduce by a standard property of measures that $\mu(E_n)&lt;\infty$ for each $n$. As a consequence, $\int |f|^p d\mu=\sum_{n=1}^{\infty}\int_{E_n}|f|^p d\mu$ Now, observe that for each $n$, $n^p\chi_{E_n} \geq |f|^p\chi_{E_n}\geq (n-1)^p\chi_{E_n}$ Upon taking integrals, we obtain: $ \sum_{n=1}^{\infty}(n-1)^p \mu(E_n)\leq\int |f|^p d\mu \leq \sum_{n=1}^{\infty} n^p\mu(E_n)$ So, if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$ then $f\in \mathcal{L}^p$. Now for the reverse, not that if $\mu(E_n)=0$ for each $n$ then we are done. If there is some $n$ such that $\mu(E_n)=0$  if $f\in \mathcal{L}^p$, then $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)&lt;\infty$. Now the question is how do we determine whether a series coneverges given another one converges? The answer is the $\textbf{limit comparison test}$: Theorem: Let $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ be two series such that $a_n\geq 0$ and $b_n&gt;0 0$ for all $n$.  Put $c= \lim_{n\rightarrow \infty}\frac{a_n}{b_n}$.  If $0&lt;c&lt;\infty$ then either both series converge or both series diverge.  To invoke this theorem notice that if for all $n$, $\mu(E_n)=0$ then we are immediately done. So, we may assume that for some n , $\mu(E_n)&gt;0$. If for all $n$, this is the case, then we are immediately done. If only finitely many satisfy this property, then we may disregard the ones with 0 measure. Thus, we may assume for each $n$, $\mu(E_n)&gt;0$. In which case, the limit comparison test above allows us to deduce that $c=1$. Since $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)$ converges, so must $\sum_{n=1}^{\infty}n^p \mu(E_n)$.
43906
I have a question that I need help with getting started (possibly I would be back for more help).     I have a measure space $(X,A,\mu)$ that is finite, and $f \in L^{\infty}(\mu)$. Also, defined is $a_n = \int_X\,|f|^n\,d\mu$. I need to show that the limit is: $$\lim_{n\to\infty}\,\frac{a_{n+1}}{a_n} = \|f\|_{\infty} .$$   I am stuck on getting started, anybody have any suggestions?  thanks much
8643
Statement  Let $(X,\Sigma,\mu)$ be a measurable space and let $(\mu_n)_{n\geq 1}$ be a sequence of measure in this space. Suppose that this is a monotone increasing sequence, in the sense that $\mu_n(E)\leq \mu_{n+1}(E)$ for all $E \in \Sigma$, and for all $n$. If we define, for every $E \in \Sigma$  $\space$  $\mu(E)=\lim_{n \to \infty} \mu_n(E)$,   prove that $\mu$ is a measure.  The attempt at a solution  I am having problems trying to prove the countable additivity property, which is, if $\{E_i\}_{i \in \mathbb N}$ is a countable collection of pairwise disjoint sets in $\Sigma$, then  $\mu(\bigcup_{i \in \mathbb N} E_i)=\sum_{i \in \mathbb N} \mu(E_i)$  In this particular case, the property is satisfied if and only if  $\lim_{n \to \infty} \mu_n(\bigcup_{i \in \mathbb N} E_i)=\sum_{i \in \mathbb N}\lim_{n \to \infty}\mu_n(E_i)$  Each $\mu_n$ is a measure, so  $\lim_{n \to \infty} \mu_n(\bigcup_{i \in \mathbb N} E_i)=\lim_{n \to \infty} \sum_{i \in \mathbb N} \mu_n(E_i)$.   How could I go on from here? I know I must use the fact that $(\mu_n)_{n \in \mathbb N}$ is a monotone increasing sequence, I would appreciate any help.
16651
Wikipedia claims, if $\sigma$-finite the Dominated convergence theorem is still true when pointwise convergence is replaced by convergence in measure, does anyone know where to find a proof of this? Many thanks!  Statement of the theorem:  Let $\mu$ be $\sigma$-finite, $|f_n|\leq g$ and $f_n\rightarrow f$ in measure, then we must have  $\int f_n \rightarrow \int f$ and $\int|f_n-f| \rightarrow 0$
121382
Can somebody provide me with a sequence of (real-valued) functions, say on $[0,1]$ with the Lebesgue measure, such that the sequence converges in probability, or maybe in $\Vert \cdot \Vert _{L^2}$, but does not converge almost surely?
50405
I'm trying to prove the following statements in Folland's book.  Let $(X,\mathcal{M},\mu)$ be a measure space. If $f_n\to f$ in measure and $g_n\to g$ in measure, then $f_n+g_n\to f+g$ in measure and $f_ng_n\to fg$ in measure (in this last case, assume $\mu(X)&lt;\infty$).  The first one is already done, but the second one is giving me trouble.   My attempt: We can take subsequences $f_{n_k}$ and $g_{n_k}$ such that $f_{n_k}\to f$ a.e. and $g_{n_k}\to g$ a.e. In this case, we know that $f_{n_k}g_{n_k}\to fg$ a.e. Note that $\mu(X)&lt;\infty$, so we can use Egoroff's theorem to conclude that $f_{n_k}g_{n_k}\to fg$ almost uniformly. Finally, almost uniform convergence implies convergence in measure, therefore $f_{n_k}g_{n_k}\to fg$ in measure.   The problem is that I just have convergence in measure for some subsequence of $f_ng_n$ and I need convergence in measure for the whole sequence. Also I don't know if using Egoroff's theorem is unnecessary, if this is the case, I would appreciate to see some more elementary prove.  Detail: We say that $f_n\to f$ almost uniformly if for every $\varepsilon &gt; 0$ there is some $E\subset X$ measurable, such that $\mu(E) &lt;\varepsilon$ and $f_n\to f$ uniformly on $E^c$.  Thank you!
122905
Let $f:[0,1] \to \mathbb R$ be a continuous function and the sequences $(a_n)_n,(b_n)_n$ s.t. $$\lim_{n\to \infty}  \int_0^1 |f(x)-a_nx-b_n| dx=0.$$ Prove that $(a_n)_n,(b_n)_n$ are convergent.   I know that $$\left|\int_0^1f(x)dx\right| \le \int_0^1|f(x)|dx.$$ Can somebody help me, please?
79302
Given a sequence $(\mu_n)_{n\in \mathbb N}$ of finite measures on the measurable space $(\Omega, \mathcal A)$ such that for every $A \in \mathcal A$ the limit   $$\mu(A) = \lim_{n\to \infty} \mu_n(A)$$   exists. I want to show that $\mu$ is a measure on $\mathcal A$.   What I managed to figure out:   $\mu$ is monotone, additive and - if $\lim_n \mu_n(\Omega)$ is supposed to be taken in $\mathbb R$ - then $\mu$ is also finite (I'll assume this). Also $\mu(\varnothing) = 0$. So we can wlog assume that $\mu(\Omega) = 1$ and that $\mu_n(\Omega) \le 2$ for all $n$.   Now what remains to be shown is that $\mu$ is $\sigma$-additive, or equivalently that for $A_n\downarrow \varnothing$, we have $\mu(A_n) \to 0$ (since $\mu$ is finite).  All attempts at proving this have been futile so far. I don't seem to see the right approach.  If possible, I would like to only receive a hint rather than a full answer. But of course, I'd be quite happy with a full answer, too; if a good hint is hard to find!  Thanks a lot in advance for your help! =)