Suppose $\mu$ is a measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. Then show that, for $\mu$-almost every $x$ $$  \int_\mathbb{R} \frac{1}{(x-t)^2}\,\mathrm{d}\mu(t) = \infty $$  I am not sure where to start. Any hints are welcome.
120937
Prove, or give a counter example: Let $\mu$ be a finite positive borel measure on $\mathbb{R}$. Then $\int (x-y)^{-2} d \mu (y) = \infty $ almost everywhere on $\mu$ (for the selection of x's).  This is a question I had in an exam, and the answer is supposed to be presented in less than 30 words, so there must be something quite simple I'm missing.
36431
Ran across this problem and need some help.   Let $E$ be a measurable subset of the real numbers and suppose that for any open interval $I$ one has $m(E\cap I)\le \frac{1}{2}m(I)$, where $m$ is the Lebesgue measure. Prove that $m(E)=0$.
107189
I am struggling to solve an exercise in my measure theory book and any help for solving it would be appreciated:  Let $(\Omega,\mathcal{A},\mu)$ be a measure space and let $f:\Omega  \to \mathbb{R}$ be measurable. Find a function $g:\Omega  \to \mathbb{R}$ which is equal to $f$ almost everywhere but is not measurable.
56240
I have been struggling with the following for quite some time now. If anyone can give me some help, it will be much appreciated:  Let $f$ bounded, measurable and $E$ be a set of finite measure. Let $A \subset E$ be measurable. Prove that:  $\displaystyle \int_{A} f = \int_{E} f \chi_A $
113366
I am having trouble with the following problem:      $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$:   $$\int_{[0,a]}f\,dm=0.$$   Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).    I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.  My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.  Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.  Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions?
71467
Assume $f:\mathbb R\rightarrow  {\mathbb R} $ is Lebesgue integrable, prove $\sum_{n\in\mathbb N } f(x+n)$ converges for almost everywhere x. This is a question discussed with friends. Thanks
111937
I'm doing exercise in Real Analysis of Folland, and got stuck on this problem. I try to use Fatou lemma but can't come to the conclusion. Can anyone help me. I really appreciate.          Consider a measure space $(X, M, \mu)$.Suppose $\{f_{n}\} \in L^{+}, f_{n} \rightarrow f$ pointwise and $\int f = \lim \int f_{n} &lt; \infty$. Then $\int_{E} f = \lim \int_{E} f_{n}$ for all $E \in M$. However, this need not be true if $\int f = \lim \int f_{n} = \infty$      Thanks so much.
46567
Let $(X, \mathbb{X}, \mu)$ be a finite measure space and let $1 \leq p&lt; \infty$. Let $\phi$ be continuous on $\mathbb{R}$ to $\mathbb{R}$ and satisfy the condition: $(*)$ there esists $K&gt;0$ such that $|\phi (t)|&lt;K|t|$ for $t \geq K$. Show that $\phi \circ f$ belongs to $L_p$ for each $f \in L_p$ . Conversely , if $\phi$ does not satisfy $(*)$, then there is a function $f$ in $L_p$ on a finite measure space such that $\phi \circ f$ does not belong to $L_p$.  Any suggestion? I tried to use $|\phi (t)|&lt;K|t|$ exchanging $t$ by $f(t)$ and I calculated $||\phi \circ f||_p$, but without success.
101607
Let  $\bigcap_{n=1}^\infty E_n=∅$ and if $\mu^*(E_n) &lt;\infty$ and $E_{n+1}  \subseteq  E_n $ then $\lim\limits_{n\mapsto \infty} \mu^*(E_n) =0 $ even if each $E_n$ is a non-measurable set, where $\mu^*$ is outer measure. Proof sketch please?
47564
Let $\mu$ be a probability measure on a metric space $M$ (with the Borel $\sigma$-algebra). If $\mu(A)\in \{0,1\}$ for all measurable set $A\subset M$, then:     Is it true that $\mu$ is a Dirac measure?   I think the answer should be negative, but I do not know of a counterexample. There is an answer  for the general case where we don't know anything about the topology (metrizable in our case). There you can see also that the result holds for all Polish spaces.
70465
I'm trying to figure out if the following is true or false:  Let $\mu$ and $\nu$ be finite measures on Borel subsets of $[0,1]$,and $\nu \ll \mu$ ($\nu$ is absolutely cts with respect to $\mu$). Does there exist always a $C \in \mathbb{R}$ such that $\forall A \subseteq [0,1]$ that is Borel, we have $\nu(A)$ $\lt$ $C \mu (A)$.  The Radon-Nykodym theorem seems to almost suggest that is true, if I knew it was bounded, then I could say $\nu (A) \le \int_{A} g \ d\mu \le \int_{A} C \ d\mu \le C\mu (A)$, where $C$ is the bound. However, I don't know if it is bounded and in fact, I could take an unbounded function that is still integrable on $[0,1]$, such as $\frac{1}{\sqrt x}$.  I can't even be sure if it is true or false.   Thank you!
76078
Suppose $(\mathbb{R},\Sigma(m),m)$ is our measure space, where $m$ is Lebesgue measure.  Also, suppose $f : \mathbb{R} \to [-\infty, \infty]$ is a Lebesgue measurable function.  The problem:     Prove that $f$ is equal almost everywhere to a Borel measurable function.   My attempt:  We only have to prove this assertion for a non-negative Lebesgue measurable function $f$ because the result will follow for all Lebesgue measurable functions.  Furthermore, since any Lebesgue measurable function can be approximated by a sequence of non-negative, monotonically increasing, Lebesgue measurable simple functions $s_{n}$, we only need to prove the claim for an arbitrary Lebesgue measurable simple function.  So, let $s: \mathbb{R} \to [-\infty, \infty] $ be a Lebesgue measurable simple function.  We can write $s$ canonically as $$ s(x) = \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(x)$$ where $\alpha_{i} \in [-\infty, \infty]$, $\bigcup \limits_{i = 1}^{n} A_{i} = \mathbb{R}$, and $A_{i} \cap A_{j} = \emptyset$ if $i \neq j$.  For each $i$, since $A_{i}$ is Lebesgue measurable, we can find a Borel set $B_{i}$ such that $A_{i} \subseteq B_{i}$ and $m(B_{i} \setminus A_{i}) = 0$.  Clearly, this implies that $\bigcup \limits_{i = 1}^{n} B_{i} = \mathbb{R}$.  Now we just need the $B_{i}$'s to be pairwise disjoint, with each $B_{i}$ still retaining $A_{i}$.  To make them pairwise disjoint, I constructed the following sets:  $\tilde{B_{i}} = [B_{i} \setminus (\bigcup \limits_{j \neq i} B_{j})] \cup A_{i}$.  This construction gives us that the $\tilde{B_{i}}$'s are pairwise disjoint (I think....) and $A_{i} \subseteq \tilde{B_{i}}$.  But I don't know that $\tilde{B_{i}}$ is necessarily still a Borel set. :( :(  Am I approaching this problem all wrong?
37398
Can I ask a homework question here?  Let $f$ be measurable and non-negative in $\mathbb R^d.$  Using Fubini's theorem, show that for $1 \leq p \lt \infty,$  $$\lVert f\rVert^p_p = \int^{\infty}_{0}pt^{p-1}\lambda(\{x:f(x)\gt t\}) \ dt.$$
115389
Suppose that $(X,\mathcal{E},\mu)$ is a non-atomic finite measure space (i.e. for every $E \in \mathcal{E}$ with $\mu(E)&gt;0$ there exists $F \subset E$ measurable such that $0&lt;\mu(F) &lt;\mu(E)$.)  a) Prove that for every $ \varepsilon &gt;0$ there is a finite partition of $X$ in measurable subsets $X_1,..,X_n$ such that $\mu(X_i)\leq \varepsilon$.  b) Prove that for every $\alpha \in [0,\mu(X)]$ there exists $E \in \mathcal{E}$ with $\mu(E)=\alpha$.    I guess that a) is given to prove b) more easily. I have the following idea of solution for a) (inspired from , which proves b)  )  Denote  $$ \Gamma = \{ (X_1,..,X_n) : n &gt;&gt; \mu(X)/\varepsilon, X_i \text{ are disjoint }, \mu(X_i)\leq \varepsilon \}$$ ordered by componentwise inclusion. Totally ordered parts $(Y_\alpha)$ of $\Gamma$ have an upper bound element $(\bigcup Y_\alpha^i)_{i=1}^n$ which is still in $\Gamma$.   By Zorn's lemma $\Gamma$ has maximal elements. If a maximal $(X_1..X_n)$ element is not a partition, then we can replace it with something better, of the form $(X_1,..,X_n\cup A)$ where $A \subset X \setminus (X_1 \cup..\cup X_n)$ and $\mu(A)&gt;0$.     My questions are:  1) Is my solution of a) correct? I feel that the part with the upper bound element may not work, since there might be noncountable unions.  2) What is a simpler solution of a)?
55055
I'm stuck with some problem of my Integral Calculation in Several Variables course. The problem goes like this:     Let $A\subset \mathbb{R}$ be a measurable set with $m(A)&lt;\infty$, and $f:A\longrightarrow [0,\infty)$ a Lebesgue-measurable function. Prove that:   $$f\in L^1(A)\Longleftrightarrow \sum_{n}^{\infty}m(\{ x\in A : f(x)\geq n \}) &lt; \infty.$$   The notation I used is:   $m$ as the Lebesgue measure function $L^1(A)=\{ f:A\rightarrow \mathbb{\overline{R}} : \int_{A}|f|\,\mathrm{d}m&lt;+\infty \}$   I've started defining the set $A=f^{-1}([0,\infty))$ as a numerable sum of disjoint measurable sets (because it's said it's measurable) $\sum^{\infty}_{k=0}\cup I_k$, being each $I_k$ the real interval $[k,k+1)$. I imagine I should come to some conclusion like that any unbounded (upper bound) sets of $f(x)$ with $(x\in A)$ have measure $0$.
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!
106738
Given a measure $\mu$ on $(X,M)$, define $\mu_0$ on $M$ by $$\mu_0(E) = \sup\{\mu(F): F\subset E \ \text{and} \ \mu(F) &lt; \infty\}$$ a.) $\mu_0$ is a semifinite measure. It is called the semifinite part of $\mu$. b.) If $\mu$ is semifinite, then $\mu = \mu_0$ (Use Exercise 14) c.) There is a measure $\nu$ on $M$ (in general, not unique) which assumes only the values of $0$ and $\infty$ such that $\mu = \mu_0 + \nu$.  The proof of Exercise 14 can be found  For a.) I believe we need to show that $\mu_0(E) = \infty$. Perhaps we can do this by proof of contradiction, similarly to what the proof in the link above does (actually it seems identical). For b.) I am not sure how we can show $\mu = \mu_0$ I think maybe we have to show by construction that $\mu \leq \mu_0$ and $\mu \geq \mu_0$ For c.) I haven't any idea for. Any suggestions is greatly appreciated.