Prove that is $E$ is measurable and $0&lt;m(E)&lt;\infty$, then for any $0&lt;\epsilon&lt;1$ there is an open interval $I$ such that $m(E \cap I)&gt;(1-\epsilon)m(I)$  How would I solve this without using The Lebesgue density theorem
121200
(Not mentioned in title but $\epsilon$ is a number greater than $0$ and $E$ a Lebesgue measurable subset of $\mathbb{R}^n$.) I know a question equivalent to this one has been asked (). But it was not answered. Also my thinking is a bit different. So I would love to get some help on this problem. So far I've got:  $$\frac{\lambda(E \cap I)}{\lambda(I)} &gt; 1- \epsilon \iff \frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)} &lt; \epsilon$$  And also, a theorem states that, for all $\epsilon$ there exists a closed set $F\subset E$ such that $\lambda(E\setminus F) &lt; \epsilon$ so it would sufice to prove  that there exists an interval $I$ such that $$\frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)}&lt;\lambda(E\setminus F).$$ I'm not sure if this helps me much but any help is appreciated. What would you do?
99739
Let $f$ be nonnegative and measurable. Show that $\int f =0$ implies $f=0$ almost everywhere (except possibly on a set of measure zero).  My thoughts. Take a set E with $m(E) \neq 0$. Suppose $f \neq 0$ on E. That means $\int_E f \neq 0$. Since $f$ is nonnegative this implies $\int f &gt;0$. That contradicts the given information, so $f$ must be 0 almost everywhere.
18402
Let $f$ integrable and let $E_\alpha=\{x\mid |f(x)|&gt;\alpha\}$ for $\alpha\geq 0$. Show that $$\int\left|f(x)\right|dx=\int_0^\infty m(E_\alpha)d\alpha$$ where $m$ is the Lebesgue measure.   By hypothesis, $f$ is measurable, therefore there exists a sequence of simple function such that $\varphi_n\nearrow |f|.$ Moreover, $E_\alpha^n\nearrow E^\alpha$. By the continuity of the measure, $m(E_\alpha^n)\to m(E_\alpha)$ for all $\alpha\geq 0$. By the convergence monotone theorem, $$\int_0^\infty  m(E_\alpha)d\alpha=\lim_{n\to\infty }\int_0^\infty  m(E_\alpha^n)\underset{(*)}{=}\lim_{n\to\infty }\int\varphi_n(x)dx=\int\left|f(x)\right|dx.$$   I don't understand the equality $(*)$, why $m(E^n_\alpha)=\varphi_n(x)$ ?
61111
Let $f:[0,1] \to \mathbb{R}$ be Lebesgue measurable with $f &gt; 0$ a.e. Suppose that $\{E_n\}$ is a sequence of measurable sets in $[0,1]$ with the property that $\displaystyle \lim_{n \to \infty}\int_{E_n} f \,dm = 0$. Prove that $\displaystyle \lim_{n \to \infty} m(E_n) = 0$.   This question is from an old analysis qual I am studying. So far I have tried a proof by contradiction: if $\displaystyle \lim_{n \to \infty} m(E_n) \neq 0$, then there is an $\epsilon &gt; 0$ and a subsequence $\{n_k\}$ so that $m(E_{n_k}) \ge \epsilon$ for all $k$. I am trying to somehow use this subsequence and show $\displaystyle \lim_{k \to \infty}\int_{E_{n_k}} f \,dm \neq 0$, which would give me a contradiction.  Another fact I know from my measure theory course is that, for meausurable $E \subseteq [0,1]$, the map $\displaystyle \nu(E) = \int_E f \,dm$ defines a measure on the Lebesgue measurable subsets of $[0,1]$. Will this fact be useful to me?
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.
58485
I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem.       Let $f:\mathbb{R}\to[0,\infty)$ be measurable and $f\in L^1$. Show that $F(x)=\int_{-\infty}^x f$ is continuous.   I know is isn't much but the only thing a could think of is that given $x, y \in \mathbb{R}$ with $x &lt; y$ we note that $F(x) \leq F(y)$, i.e. $F$ is increasing. So maybe we can apply one of the convergence theorems of Lebesgue integration theory.    I was also wondering if this problem can be solved using only Riemann integration theory.
70674
I am trying to show that for nonnegative $f$ on $\mathbb{R}$ if $||f||_1&lt;\infty$, we have  $$ \lim_{p\uparrow \infty}||f||_p=||f||_\infty $$ I have tried to fiddle around with  $$ \left(\int_\mathbb{R}|f|^p|f|^{1-p} \right)^{1/p}&lt;\infty $$ but haven't found a nice way to get an estimate on $\int_E|f|^p$ from this. Honestly I am pretty uncomfortable with non finite measure spaces.   Any hints would be awesome! Any references for problems similar to this would also be awesome.  edit: this is not a duplicate of the linked question as 1) my measure space is not finite, 2) my function is in $L^1$ and not necessarily in $L^\infty$
22587
Let $\Omega$ be a measure space with measure $\mu$. (Here, a measure is only meant to be countable additive!)  Consider a subset $A\in\Sigma$.  Then according to the  it is an atom if: $$(1)\quad\forall E\subseteq A:\quad\mu(E)&lt;\mu(A)\implies\mu(E)=0\quad(\mu(A)&gt;0)$$ and according to the  it is an atom if: $$(2)\quad\forall E\in\Sigma:\quad\mu(E\cap A)=0\lor\mu(E^c\cap A)=0\quad(\mu(A)&gt;0)$$  Now, these definitions agree for the really trivial cases: $$\mu\equiv0$$ $$\mu\equiv\infty$$ but they differ for the less trivial case: $$\quad\mu(E\neq\varnothing):=\infty,\,\mu(\varnothing):=0$$ namely the atoms are w.r.t. (1) all nonempty subsets whereas w.r.t. (2) only the singletons.  Excluding this pathological case, are the definitions equivalent? (More precisely, assume there exists a measurable subset $0&lt;\mu(F)&lt;\infty$.)  So far I checked that: $$(2)\implies(1):\quad\mu(E)&lt;\mu(A)\implies\mu(A\setminus E)&gt;0\implies\mu(E)=0$$ but what about the other direction?
79930
Why do we have the following equality$$\int_{-\infty}^\infty |g(x)|\,dx = \int_0^\infty \mu(\{x : |g(x)| \ge t\})\,dt,$$where $\mu$ is Lebesgue measure?
116357
If $X$ is measurable space, $f : X\to [-\infty,+\infty]$, function such that $f^{-1}([-\infty,r))$ measurable for all rational numbers $r$,prove than $f$ is measurable  I know that $f$ is measurable iff set $\{x\in X : f(x)&lt;a\}$ is measurable for all $a\in \mathbb{R}$ but for $a\in\mathbb{Q}$ I don't know how to proceed.
40284
Let $E$ be a measurable subset of $\mathbb{R}$ and let $F_n :E \to \mathbb{ R} $ be a sequence of measurable functions. Let $$E^*= \{ x\in E : F_n  \text{ is convergent } \}$$ Is the set $E^*$ measurable?    To solve this, I need to understand the relation between the set $E^*$ and the functions $F_n$. They are strongly related, but how can I express this relation in terms of sets and measure?
122265
Let $A\subset\mathbb{R}$ a measurable where $0&lt;m(A)&lt;\infty$. Show that exists for each $0&lt;\alpha&lt;1$ an interval $I$ such that  $$ \frac{m(A\cap I)}{m(I)}&gt;\alpha. $$  MY ATTEMPT: Following a hint a give that:  Let $\varepsilon&gt;0$, exists $G$, a open set such that $m(A)\leq m(G)&lt;m(A)+\varepsilon$. As $G$ is open, we can write as the disjoint sum of open intervals $\dot{\bigcup}_n I_n = G$.So, $$ m(A)\leq m(G)=m\left(\sum \dot{\bigcup_n} I_n \right)\leq \sum m\left(I_n \right)&lt;m(A)+\varepsilon $$ Suppose that $\varepsilon=(\alpha^{-1}-1)m(A)$:  $$ m(A)+\varepsilon=m(A)+m(A)(\alpha^{-1}-1)=m(A)(1+\alpha^{-1}-1)=m(A)\alpha^{-1} $$ and $$ \alpha\sum m(I_n)&lt;m(A)=\sum m(A\cap I_n) $$ $$ \Rightarrow \alpha&lt;\frac{\sum m(A\cap I_n)}{\sum m(I_n)}  $$  But I have a insecurity with that:   (1) I can write that: $m(A)=\sum m(A\cap I_n)$?   (2) How to get the result with this: $$ \alpha&lt;\frac{\sum m(A\cap I_n)}{\sum m(I_n)}  $$
111317
Below is an attempt at a proof of the following problem. Any feedback would be greatly appreciated. Thx!  Let $E$ be a set and $O_n = \{x: d(x, E) &lt; \frac{1}{n}\}$. Show  If $E$ is compact then $m(E) = lim_{n \rightarrow \infty} m(O_n)$.  This is false for $E$ closed and unbounded or $E$ open and bounded.   Note that all sets are real and measurable refers to Lebesgue measurable.  Suppose $E$ is compact. Then $m(E) &lt; \infty$. Also, since each $O_n$ is an open set of $\mathbb{R}^d$, $m(O_n) &lt; \infty$. If either of the following is true, we have that $lim_{n \rightarrow \infty}m(O_n)=m(E)$: $O_n \nearrow E$ $O_n \searrow E$ Now  $\bigcap_{n=1}^{\infty} O_n = \{ x:d(x,E) = 0\}$. That is, $x \in \bigcap_{n=1}^{\infty}O_n \iff \exists n&gt; \frac{1}{\epsilon},\ n \in \mathbb{N} \iff x\in \{x \mid d(x,E) =0\}$. Then we have that $\bigcap_{n=1}^{\infty}O_n = E$ and $O_n \searrow E$. Thus $ \ m(E) = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is closed and unbounded. Let $E = \{(x,y): y = 2 \}$ and $O_n = \{(x,y): d(x,E) &lt; \frac{1}{n}\}$. Since $E$ is a line in $\mathbb{R}^2$, $m(E) = 0$. Also, since the measure of a rectangle in $\mathbb{R}^d$ is its volume, $m(O_n) = \left| O_n \right| = \infty$, since the rectangle $O_n$ is unbounded. It is therefore apparent that $m(E) \not = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is open and bounded. Let $E=(0,1)$ and $O_n = \{x \mid d(x,E) &lt; \frac{1}{n}\}$. $\mathbb{R}-E$ is closed and $O_n \in \mathbb{R}-E$. Since $\mathbb{R}-E$ contains all its limit points $lim_{n \rightarrow \infty}O_n = O \in \mathbb{R}-E$. Thus $lim_{n \rightarrow \infty} m(O_n) \neq m(E)$.
94597
I am currently working on a proof that requires me to show that an open ball $B_{\epsilon}(x)$ has nonzero measure. I currently have the following proof in my book:  "The closed interval $[a,b]$ is not of measure zero."  Hence, if I take the contrapositive, does it follow that "if an interval doesn't have measure zero, the interval is open"? Is there a way for me to prove that the open interval on the line $\mathbb{R}$ is of measure not zero? Thanks!
106583
Let $(X,\mathcal{M},\mu)$ be a measure space and $f$ be a nonnegative measurable function on $X$. Let $1\le p&lt;\infty$. Show that, the function $|f|^p$ is integrable with respect to $\mu$ precisely when the function $$t\mapsto t^{p-1}\mu({x:|f(x)&gt;t})$$ is integrable on $[0,\infty)$ with respect to Lebesgue measure. In addition prove that    $$\int_X |f|^p=p\int^{\infty}_{0} t^{p-1}\mu({x: |f(x)&gt;t}) dt\,$$   I solved this problem for the case when space is sigma finite but don't know how to solve for general case.
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.
3388
Let $X\subset \mathbb{R}$ Lebesgue measurable, $|X|&lt;2^{\aleph_0}$, is it true that $X$ is null?  Of course I am not assuming the Continuum Hypothesis.   EDIT: It might be helpful to know that all Borel measurable sets have cardinality either $\aleph_0$ or $2^{\aleph_0}$. Then a measurable set of cardinality strictly between those two must be Lebesgue but not Borel measurable.