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.
75224
Let $({\Bbb R},{\mathcal A},\mu)$ be the measure space where ${\mu}$ is the Lebesgue measure.  Assume that $\int_{\Bbb R}|f|^pd\mu&lt;\infty$ ($p\geq1$). There is an exercise for proving that       $$\int_{[0,\infty)}pt^{p-1}\mu(\{x:|f(x)|\geq t\})d\mu(t)=\int_{[0,\infty)}\mu(\{x:|f(x)|^p\geq s\})d\mu(s).$$   If one formally let $s=t^p$, then $ds=pt^{p-1}dt$ and one would immediately get the equality above. However, how can I justify it rigorously in the sense of Lebesgue integration?    [Edited] What puzzles me is that I have the "change of variable" theorem for Riemann integration, but all the integrations here are in the Lebesgue sense. Unless one has shown that the integrands are also Riemann integrable, how can I directly use "change of variable"?
17835
Let $(X,B,\mu)$ be a complete measure space,Show that $$\lim _{q \rightarrow \infty}\|f\|_{q}=\|f\|_{\infty}, \quad \forall f \in \bigcup_{p} \bigcap_{p \leqslant q&lt;\infty} L^{q}$$ So,$\lim _{q \rightarrow \infty}\|f\|_{q}$ , $\|f\|_{\infty}$ are equal-norm with space $ L^{\infty} \cap\left(\bigcup_{p} \bigcap_{p \leqslant q} L^{q}\right)$. Case 1: $m(X)&lt;\infty $.It's easy to prove that. Case 2: $m(X)=\infty $. I have no idea about it,And I started to doubt the correctness of this conclusion. Can somebody give me a hint for this problem or just give an example to prove that this is a wrong conclusion when $m(X)=\infty $. Thanks in advance.
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!
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 $
25170
In a space with measure $1$, $||f||_p$ is an increasing function with respect of $p$. To show that $\lim_{p \rightarrow \infty} ||f||_p=||f||_{\infty}$ we have to show that  $||f||_{\infty}$ is the supremum, right??   To show that, we assume that $||f||_{\infty}-\epsilon$ is the supremum.  From the essential supremum we have that $m(\{|f|&gt;||f||_{\infty}-\epsilon\})=0$.  So, we have to show that $m(\{|f|&gt;||f||_{\infty}-\epsilon\})&gt;0$.   Let  $A=\{|f|&gt;||f||_{\infty}-\epsilon\}$.   We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$.   $\int_A |f|^p &gt;\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$   So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;||f||_p \leq ||f||_{\infty}$  How could we continue to show that $m(A)&gt;0$??   EDIT:   Is it as followed??  We have that $0&lt;||f||_{\infty}-\epsilon&lt;||f||_{\infty}$ for some $\epsilon&gt;0$.  $||f||_{\infty}$ is the essential supremum. So, from the definition we have that $m\left ( \{|f(x)&gt;||f||_{\infty}-\epsilon\} \right )&gt;0$.   Let  $A=\{|f|&gt;||f||_{\infty}-\epsilon\}$.   We have that $\int_A |f|^p \leq \int |f|^p \leq ||f||_{\infty}^p$.   $\int_A |f|^p &gt;\int_A (||f||_{\infty}-\epsilon)^p=(||f||_{\infty}-\epsilon)^p m(A)$   So, $m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;||f||_p \leq ||f||_{\infty}$  Taking the limit $p \rightarrow +\infty$ we have the following:  $$\lim_{p \rightarrow +\infty}m(A)^{1/p} (||f||_{\infty}-\epsilon)&lt;\lim_{p \rightarrow +\infty} ||f||_p \overset{ m(A)&gt;0 \Rightarrow \lim_{p \rightarrow +\infty}m(A)^{1/p}=1}{\Longrightarrow} \\ ||f||_{\infty}-\epsilon&lt;\lim_{p \rightarrow +\infty} ||f||_p$$   Is this correct?? How do we conclude that $\lim_{p \rightarrow +\infty} ||f||_p=||f||_{\infty}$ ??
44926
Let $(X, \mathcal B, m)$ be a measure space.  For $1 \leq p &lt; q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample in the case the condition is not satisfied?
64748
Recall that $\|f\|_\infty = \inf\{M\ |\ |f(x)|\leq M \text{ for almost every } x \in X\}$, for every function $f$ defined on $X$. Show that, if $\|f\|_\infty$ is finite, then $\|f\|_\infty$ is also the smallest of all numbers of the form $\sup\{|g(x)| : x \in X \}$, where $f = g$ almost everywhere with respect to the Lesbegue measure.   I am trying to understand how $L^{\infty}$ is defined. Could someone help me see how to prove the above fact?
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?
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 $\sum_{n=1}^{\infty}n^p \mu(E_n)$.
34130
I want to prove that if I omit the fact that $\mu (X) &lt; \infty$ in Egorov theorem and place instead that our functions $|f_n| &lt;g$ and $g$ is integrable, we still get the result of Egorov's theorem.  Fix $m$ a natural number.  I took $ w_{n} = |f_n-f|$ and thus by DCT $\int |f_{n} - f|$ goes to zero.  Then I took $\bigcup_n {( w_{n} \geq 1/m)}$. I need its measure to be finite. Its measure is less than the sum of the measures of each $ w_n\geq 1/m)$ varying $n$, and by Tchebychev, this is less than $m\int|f_{n} - f|$.  But I got stuck here.  Any help is appreciated. Thanks!
17300
Let $X$ be the interval $[0,1]$ with Lebesgue measure. Is there a function $f\in L^p(X)$ for all $p\in[1,\infty)$ that is not $\in L^\infty(X)$? If so, what is an example?   Motivation: In a course on measure theory this fall, I've learned proofs that $L^p(X)\supset L^q(X)$ if $q&gt;p$ and that if $f\in L^\infty(X)$, then $\|f\|_\infty = \lim \limits_{p\to\infty} \|f\|_p$.  This prompted me to wonder if $L^\infty(X) = \bigcap _p L^p(X)$. A classmate gave me a general theoretical reason to believe the contrary: $L^p(X)$ is a  for $1 \lt p \lt \infty$ but not for $p=1,\infty$; but intersections of reflexive spaces are reflexive. This logic seems sound to me; but it implies the containment $L^\infty(X) \subset \bigcap_p L^p(X)$ is strict.  If so, there must be a function that is $L^p$ for all $p$ but not a.e. bounded. What is it?
114628
From Donald Cohn's Measure Theory, section 2.4, exercise 10.  Let $(X, A, \mu)$ be a measure space, and let $f$ and $f_1, f_2, \dots$ be non-negative functions that belong to $L^1(X, A, \mu, R)$ and satisfy (i) $\{f_n\}_n$ converges to $f$ almost everywhere; (ii) $\int fd\mu = \lim_n\int f_nd\mu$. Show that $\lim_n\int |f_n - f|d\mu = 0$.  I let $f_n = 1/n$ on $[n, n+1]$ and $0$ elsewhere, and let $f=0$, then all the convergence theorems in that section (dominated convergence, and monotone convergence) failed.
41714
Problem:  Find a non-negative function $f$ on $[0,1]$ such that $$\lim_{t\to\infty} t\cdot m(\{x : f(x) \geq t\}) = 0,$$ but $f$ is not integrable, where $m$ is Lebesgue measure.  My Attempt:  Let $f(x) = \frac{\chi_{(0,1]}}{\sqrt{x}}$. Then,  \begin{align*} \lim_{t\to\infty} t\cdot m(\{x \in [0,1]: f(x) \geq t\}) &amp;= \lim_{t\to\infty} t \cdot m(\{x\in (0,1]: 1/\sqrt{x} \geq t\})\\ &amp;=\lim_{t\to\infty} t \cdot m(\{x\in (0,1]: x \leq (1/t^2)\})\\ &amp;= \lim_{t\to\infty} t \cdot m((0, (1/t^2)))\\ &amp;= \lim_{t\to\infty} \frac{t}{t^2} = 0. \end{align*}  However, $f(x)$ is integrable over this interval. I have tried functions looking like $f(x) = 1/x^p$ but I cannot find any that will work here. There is a hint that says there is a monotonic function that fits this description.    Also, does anyone know of a list of non-Lebesgue integrable functions on $[0,1]$? I feel as though I could use this for many counterexamples if one were to exist. Thanks!
36954
In what follows I'm only considering positive real valued functions.  Everywhere I look about the definition of the Lebesgue integral it is required to consider a measurable function. Why do we not define the integral for non-measurable functions? From what I see we require measurablility of the simple functions that approximate f, not f itself. The definition I'm considering is given a measure space $X$ with measure $\mu$ and a measurable function $f$ we define  $$ \int_E f \, \mathrm{d}\mu = \sup_{s \in S} \int_X s \,\mathrm{d}\mu $$ where $S = \{ s : X \to [0, \infty) \mid 0 \le s \le f, s \text{ is simple, measurable} \}$.  For example consider $\mathbb{R}$ with the sigma algebra $\varnothing, \mathbb{R}$ with measure $\mu$ given by $\mu(\varnothing) = 0, \mu(\mathbb{R}) = 1$ and consider $f = \chi_{[0,1]}$ then why can't we say that $$ \int_{\mathbb{R}} f \,\mathrm{d} \mu = 0 $$ (since the only measurable simple function such that $0\le s \le f$ is $s = 0$) which would follow the definition above? Is this not well defined? In general I'm struggling to see why measurable functions (other than measurable simple functions) are used.
104930
I'm working in the following exercise:  Suppose $(X, \mathcal A, \mu)$ is a finite measure space and suppose $\mathcal F$ is the set of all $\mathcal A$-measurable functions $f: X \rightarrow \mathbb R$. For $f, g \in \mathcal F$, let $$d(f, g)=\int_X\frac{|f-g|}{1+|f-g|}d\mu.$$  Show that:   $d(f, g)=0$ if and only if $f=g$ almost everywhere. $d(f, g)=d(g, f)$ $d(f, g)\leq d(f, h)+d(h, g)$ If $f_n$ is a sequence in $\mathcal F$ and if $f \in \mathcal F$ then $d(f_n, f)\rightarrow 0$ if and only if for every $\delta&gt;0$ the following holds: $$\lim_{n\rightarrow\infty}\mu(\{x \in X: |f_n(x)-f(x)|\geq\delta\})=0.$$ If $f_n$ is a Cauchy sequence on $\mathcal F$ then there exists $f \in \mathcal F$ such that $d(f_n, f)\rightarrow 0$.   i.e., I must show that this thing is a complete pseudometric space.  I have already done 1. 2. 3. 4., but I'm stuck on 5. I don't know which $f$ should the $f_n$ converge to.
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.
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.