Let $\mu$ be a semi-finite (and countably-additive) measure and $\mu(E)=\infty$. Then for any $C&gt;0$, there exists a measurable set $F$ such that $F \subseteq E$ and $C&lt;\mu(F)&lt;\infty$.  By definition of semi-finiteness, I know I am guaranteed that there will exist a measurable-set $F$ satisfying $F \subseteq E$ and $0&lt;\mu(F)&lt;\infty$. I am not sure how can I construct a set satisfying what I am required to prove. Will appreciate some hint/help. Many thanks.
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.
121971
I was wondering if someone could please give me an example of a sequence of decreasing sets where the first set has infinite Lebesgue measure; i.e., $\{B_{n}\}_{n=1}^{\infty}$ such that $m(B_{1}) = \infty$ but $m(\cap_{n=1}^{\infty} B_{n}) \neq \lim_{n \to \infty}m(B_{n})$?  thank you.
68470
I'm a newcomer in real analysis. I am learning the concept of measurable by myself using Royden's book "Real Analysis". I have a question regarding measurable sets. The following definition comes from Royden's book (page 35).  Definition: A set $E$ is said to be measureable provided for any set $A$, $$m^*(A)=m^*(A\cap E)+m^*(A\cap E^C)$$ where $m^*(\cdot)$ denotes the outer measure of a set.  To me, intuitively the above equation holds for all sets. In $\mathbb{R}$, I think a set can either be an interval or a series of isolated points (right?). It seems these kinds of sets are all measurable by the definition. Can anybody give me an example of non-measureable sets so that I can have an intuitive understanding regarding this concept? Thanks.
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.
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 $m(E)=0$.
105043
Suppose we have a premeasure $\mu$ on a space $X$ such that $\mu(X) &lt; \infty$. Prove that $E \subset X$ is Caratheodory measurable iff $ \mu^*(E)+ \mu^*(E^C) = \mu(X)$.  Going in the forward direction is fairly easy. Assuming that E is Caratheodory measurable, we can just substitute X into $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^C) $, and then we note that the outer measure and the premeasure of X themselves would have the same value.   The other direction is more difficult however. My primary idea of how to solve this part is to show that $\mu^*(A)$ and $\mu^*(A \cap E) + \mu^*(A \cap E^C) $ are both greater than and less than each other to show equality. However, I am not completely sure how to proceed with this. Can anybody provide any pointers as to how to prove the equality between these two expressions?
26388
Give an example of a measure space $(\Omega, \mathit{F})$ and a function $\mu$ on $\mathit{F}$ that is additive but not $\sigma$-additive, i.e. $\mu(\cup A_i)= \sum\mu(A_i)$ for a finite collection of disjoint $A_i$ but not for some infinite collections.   I know a measure function defined on $\sigma$-algebra is $\sigma$-additive, but I struggle finding a function that would not be additive for infinite collections. Can someone give me an example and show me why?
25964
Consider two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$, where $\mu\times\nu(X\times Y)&gt;0$. Given two measurable functions $f:X\to \mathbb{R}$ and $g:Y\to\mathbb{R}$ such that   $$f(x) = g(y)   \qquad\mu\times\nu \,\,\text{a.e,}$$  does it follow that there exists a constant $\lambda$ such that $f(x)=\lambda$ for $\mu$-a.e. $x$ and $g(y)=\lambda$ for $\nu$-a.e. $y$?  Clarification: The displayed statement means that $\mu\times\nu\big(\{(x,y):f(x)\neq g(y)\}\big)=0$.  I know that this is true when you assume $\mu$ and $\nu$ are $\sigma$-finite, by applying Fubini's theorem to $\int |f(x)-g(y)|\,d(\mu\times\nu)$. In the general case, the only first step I can think of is to find a countable set of rectangles with arbitrarily small total area where $f(x)=g(y)$ on the complement of their union. But there I'm stuck.
47378
Could someone please check if I'm saying something that make sense? I am trying to check the following:     If $f$ is a measurable function, then $\text{supp}(f)$ is measurable set?   Is it simply because, if we consider $E_1 = (0,\infty)$ and $E_2 = (-\infty,0)$, since they are open, they are measurable. So their union $\mathbb{R}-\{0\}$ is measurable as well. Since $f$ is measurable, then the preimage $f^{-1}(\mathbb{R}-\{0\})$ is measurable. But the preimage is just the support?
111083
Let $(f_n)$ be a sequence of measurable functions defined on a measurable set $E$. Is the set $A=\{x\in E|\lim_{n \to\infty} f_n(x)\ exists\ (finitely)\}$ measurable?  The following is my answer:   Define $g:A\to\mathbb{R}$ as $g(x)=\limsup f_n(x)-\liminf f_n(x) \ \forall x\in A$. Then $g(x)=0 \ \forall x\in A$ since $x\in A\implies \limsup f_n(x)=\limsup f_n(x)=\lim f_n(x)$ and $g$ is measurable as $\limsup f_n,\ \liminf f_n$ are measurable functions. Thus $g^{-1}((-1,1))=A$ is measurable as $(-1,1)$ is open.  Can somebody please tell me if this proof is alright? Thank you.
95018
Let $S$ be a bounded measurable subset of $\mathbb R$. Let $f \colon S  →  (0,\infty)$ be Lebesgue integrable. Prove that   $$\lim_{n\to\infty}\int_S\ f^{1/n} \;\mathrm{d}m = m(S)$$  Where $m(S)$ is the Lebesgue measure of $S$.  There is a hint that let $S = A \cup B$, where $A = f^{−1}((0, 1))$ and $B = f^{−1}([1,+\infty))$.   And the sequence of functions $h_n\ (n = 1, 2, 3, \dots),$ where $h_n(x) = ({f(x)})^{1/n}$.  I notice that $h_1(x)\leq h_2(x)\leq h_3(x) \leq \dots$   How can I apply monotone convergence theorem to the given hint.
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?
69809
I am reading a theorem about measurability of vector-valued functions in a note on functional analysis:     Theorem 3.6.1. If $X$ is a separable, metrizable locally convex space, $(\Omega, \Sigma, \mu)$ is a $\sigma$-finite measure space, and $f : \Omega \to X$ is weakly measurable, then $f$ is strongly measurable.   The proof begins     We can restrict ourselves to the case of finite measures.   Could somebody explain how the general case (when $(\Omega,\Sigma,\mu)$ is "$\sigma$-finite") can be deduced from the "finite" case?    For completeness, here are some definitions. Let $( \Omega, \Sigma, \mu )$ be a measure space, let $X$ be a Hausdorff locally convex space, and let $f : \Omega \to X$. We say that   $f$ is measurable if $f^{-1} (G) \in \Sigma$ for every open set $G \subset X$. $f$ is weakly measurable if $\varphi \circ f$ is measurable for every $\varphi \in X^*$. $f$ is strongly measurable if there exist simple functions $f_n : \Omega \to X$ such that $f(\omega) = \lim_{n \to \infty} f_n(\omega)$ for $\mu$-almost every $\omega \in \Omega$. (A simple function is a measurable function which takes only finitely many values.)   An image of proof, and surrounding definitions, can be seen .
87046
Let X be an uncountable set. And let $$\mathcal{A} =\{A\subset X: A \text{ is countable or } X-A \text{ is countable}\}.$$ Then it can be shown that $\mathcal{A}$ is a $\sigma$-algebra. Define $\mu:\mathcal{A}\rightarrow [0,\infty]$ by $\mu(A)=0$ if $A$ is countable or 1 otherwise. The task is to show that $\mu$ so defined is a measure on $\mathcal{A}$.  It is clear that $\mu(\emptyset)=0$. Now we want to show that $\mu$ is countably additive on disjoint sets from $\mathcal{A}$. I'm first trying to show this for two disjoint sets $A$ and $B$ in $\mathcal{A}$. I am running into a problem if I assume that both $X-A$ and $X-B$ is countable.  For if this happens then, $B\subset X-A$ implies $B$ is countable. But both $B$ and $X-B$ cannot be countable, since then $X=B\cup (X-B)$ will be countable, running contrary to the assumption that $X$ is uncountable. How should I address this case?
45022
Question: Let $E_1,E_2,...,E_n$ be measurable subsets of $[0,1]$ with $$ \sum_{k=1}^n m(E_k)  &gt; n-1.$$ Prove that $\bigcap_{k=1}^n E_k$ has positive measure.  This is one of the questions in graduate analysis past year paper. I think we need to assume that the intersection $\bigcap_{k=1}^n E_k$ is nonempty. Otherwise, the question is false. Anyway, let's assume that $\bigcap_{k=1}^n E_k \neq \emptyset.$ If $n=2,$ by inclusion-exclusion principle and assumption, $$m(E_1\cup E_2) + m(E_1 \cap E_2) = m(E_1) + m(E_2) &gt; 2.$$ Since $E_1,E_2 \subseteq [0,1],$ by monotonicity of Lebesgue measure, $$m(E_1) \leq 1, m(E_2) \leq 1.$$ By finite subadditivity of Lebesgue measure, $$m(E_1 \cup E_2) \leq m(E_1) + m(E_2) \leq 2.$$ Hence, $$m(E_1 \cap E_2) &gt; 2 - m(E_1 \cup E_2) \geq 0$$ Therefore, the intersection $\bigcap_{k=1}^n E_k$ has positive measure for $n=2.$ I try prove the general $n$ using induction, but it seems a bit long. Does there exist an efficient method to solve the question? EDIT: Actually I am looking for a direct proof instead of indirect proof or inductive proof. If a direct proof is not possible, then I will accept Marios's answer.
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.