If $f_n$ and $f$ are integrable, $f_n \to f$ a.e., and $\int |f_n| \to \int |f|$, then $\int |f_n-f| \to 0$.  How can I prove this? The limit definition proof involving $N$ and $\epsilon$ is not working for me ... I can't get a certain triangle inequality to work. But is this kind of proof the way to arrive at the result?
7520
Let $1\leq p &lt; \infty$. Suppose that    $\{f_k\} \subset L^p$ (the domain here does not necessarily have to be finite),  $f_k \to f$ almost everywhere, and  $\|f_k\|_{L^p} \to \|f\|_{L^p}$.    Why is it the case that $$\|f_k - f\|_{L^p} \to 0?$$   A statement in the other direction (i.e. $\|f_k - f\|_{L^p} \to 0 \Rightarrow \|f_k\|_{L^p} \to \|f\|_{L^p}$ ) follows pretty easily and is the one that I've seen most of the time. I'm not how to show the result above though.
48503
How to formally prove that if $\lim \limits_{n \to \infty}a_n=\infty$, then  $\lim \limits_{n \to \infty}\frac{1}{a_n}=0$.  I got confused for how to mix between convergence to infinity and having finite limit. how do I write a proof to link between two things? ( by using $\epsilon $) and not just giving an example.  my proof:   let $M= \frac{1}{\epsilon}$. there exists such a $N\in\mathbb N$ s.t for every $n&gt;N$, it is true that: $$a_n&gt;\frac{1}{\epsilon}$$  we should prove that for every $\epsilon&gt;0$ there exists such a  $N\in\mathbb N$, s.t for every $n&gt;N$ $$\left|\frac1{a_n}\right|&lt;\epsilon$$  so we take the N that satisfies the first conclusion, and that will mean for every $n&gt;N$$$\left|\frac1{a_n}\right|&lt;\epsilon$$
92992
Let $S=\{f:\mathbb{R} \to  \mathbb{R}\}$ that satisfies: $\forall f\in S$, $f'$ exists and $f'$ is continuous and $f(0)=f(1)=0$.  Please prove that $\exists \delta :\forall f\in S$ s.t.  $\int_{0}^{1} (f(x))^2dx\leq \delta \int_{0}^{1} (f'(x))^2dx$. Thanks in advance.
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.
122166
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of Riemann-integrable functions on $[a, b]$.   Assume that $(f_n)_{n\in\mathbb{N}}$ converges pointwise to $f$ on $[a, b]$ and converges in the mean to $g$ on $[a, b]$, where $f$ and $g$ are both continuous. Show that $f = g$.   ($(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$ means that $\lim_{n\to\infty} d(f_n, g) = 0$, where $d$ is the $L_2$ metric.)  I am studying real analysis now and want to prove this statement. But I don't know how to use condition "$(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$." How can I prove this?
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
78763
Let $f$ be a not decreasing, Lebesgue integrable function with finite integral over $[0,+\infty)$.  It seems obvious to me that $\lim_{b\rightarrow\infty}\int_b^{\infty}f(x)\,dx=0$ then.  But how can one very formally show it?  And assumption about monotonicity is crucial here, yes?
77431
If $f$ is bounded by $M$ on $[a, b]$ and if the restriction of $f$ to every interval $[c, b]$ where $c$ in $(a, b)$ is Riemann integrable, then $f$ is Riemann integrable and that $\int _c^b f \to \int_a^b f$ as $c \to a+$.  I am stuck with this problem.
69661
Let $f(x)$ be continuous on $[a,b]$ and suppose $\int_a^b f(x)g(x)dx = 0$ for every continuous function $g$ on $[a,b]$. Prove that $f(x)=0$ on $[a,b]$.   I understand that $f(x)$ must be zero otherwise the integral can't possibly be zero for all possible $g$. But, I am not sure how to approach this systematically. Any suggestions would be appreciated. Thanks!
30284
How to show that $\lim_{x \to +\infty}(f(x)+f'(x))=0 $ implies $\lim_{x \to +\infty} f(x)=0$?
42082
Let $f_1 : [a,b] \rightarrow \mathbb{R}$ be an integrable function.  Let's define a sequence $(f_n)$,  $ \ \ f_n : [a,b] \rightarrow \mathbb{R}$ as $f_{n+1}(x):= \int_a ^x f_n(t)dt$.'  Prove that  $\sum_{m=1} ^{\infty} f_m(x)$ is uniformly convergent on $[a,b]$.  Could you help me with that?  Thank you.
18829
I have a question:     Suppose $f$ is continuous and even on $[-a,a]$, $a&gt;0$ then prove that   $$\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$$   How can I do this? Don't know how to start.
17231
Trying to solve     $f(x)$ is uniformly continuous in the range of $[0, +\infty)$ and $\int_a^\infty f(x)dx  $ converges.   I need to prove that: $$\lim \limits_{x \to \infty} f(x) = 0$$  Would appreciate your help!
123995
I need help proving that if f is Lebesgue integrable then $$\sum_{k \in \mathbb{Z}}{ 2^k*m(|f| &gt; 2^k)}$$  converges.   I proved the opposite direction but I don't know how to put any bounds on each term, I would appreciate hints/suggestions.
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 $m(E^n_\alpha)=\varphi_n(x)$ ?
30814
Suppose $f$ is a Riemann integrable function on $[0,1]$. Prove that $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$   This is what I am thinking: Fix $n$. Then by Jensen's Inequality we have $$0\leq\left(\int_0^1x^nf(x)dx\right)^2 \leq \left(\int_0^1x^{2n}dx\right)\left(\int_0^1f^2(x)dx\right)=\left(\frac{1}{2n+1}\right)\left(\int_0^1f^2(x)dx\right).$$Thus, if $n\to\infty$ then $$0\leq \lim_{n\to \infty}\left(\int_0^1x^nf(x)dx\right)^2 \leq 0$$ and hence we get what we want. How correct (or incorrect) is this?
26735
I'm trying to prove the following   Theorem: Let $\{f_n\}_{n\in\mathbb N}\subset L^p(\Omega)$, $f_n \to f$ in $L^p(\Omega)$ ($\Omega\subset\mathbb{R}^n$ is open and bounded, $1\leq p \leq \infty$) and $f_n \to \hat{f}$ almost everywhere. Then $f=\hat{f}$ almost everywhere.  Ideas for the proof:  We should prove that $$\int (f-\hat{f}) =0$$ We can write $$\int (f-\hat{f}) = \int (f-f_n)+\int (f_n-\hat{f})$$ Then, by weak convergence, the first integral tends to $0$ (because $1\in L^\infty\subseteq L^p$).  The problem is the limit of the second integral. We can write $$ \int (f_n-\hat{f})=\int_{\Omega \cap E} (f_n-\hat{f}) + \int_{\Omega\cap E^c} (f_n-\hat{f})$$ where $E=\{x: f_n(x) \not \to \hat{f}(x)\}$. Then $|E|=0$ and the first integral vanishes.   My question is: What should I do with the second integral? Is it enough if I use the Lebesgue Dominated Convergence Theorem, since by Hölder inequality I can easily dominate $\{f_n\}$ in $L^1$?