I want to show the following converse of an (apparently) common integration theory exercise. Suppose $f:[0,1]\rightarrow\mathbb R$ is integrable, then $$\sum_{n=1}^\infty 2^n\lambda(\{\lvert f\rvert\ge 2^n\})&lt;\infty.$$ Since series convergence is annoying to prove, I thought the best way to tackle this problem would be by assuming that the series doesn't converge, then showing that $f$ can't be integrable. However, I couldn't get much from the series not converging, as the terms of the series might well go to $0$ even if the series doesn't converge. Is there something I'm missing to make this approach work? Or is a direct proof easier somehow?
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.
15494
Determine whether the series $\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}$ is convergent or divergent.  I've tried applying all the basic tests to no avail. I need to find out what the "trick" for this particular series is.
36388
Let us define the Fourier transform of the Lebesgue-summable function $f\in L_1(\mathbb{R},\mu_x)$ as $F[f](\lambda)=\int_{\mathbb{R}}f(x) e^{-i\lambda x} d\mu_x$, where $\mu_x$ is the Lebesgue linear measure. Kolmogorov-Fomin's (p. 430 ) states that, if $x\mapsto xf(x)$ is Lebesgue-summable too, then the Fourier transform is differentiable and$$\frac{d}{d\lambda}F[f](\lambda)=-i\int_{\mathbb{R}}x f(x) e^{-i\lambda x} d\mu_x$$  EDIT: I would be very grateful to the people having voted to close the question, and to anybody else, if they explained how to use the  to prove this equality, which I am not able to do. I am trying to teach myself and cannot afford university for the moment and have not got the sufficient mathematical ability to understand how that result implies this equality. I heartily thank you all!!!  Previous trial of mine which may or may not be useful to prove the given equality: In order to prove the statement to myself I had thought that the fact that if the sequence $\{\varphi_n\}\subset L_1(\mathbb{R},\mu_x)$ converges to $\varphi$ with respect to the metric of $L_1(\mathbb{R},\mu_x)$, then $F[f_n]\xrightarrow{n\to\infty} F[f]$, but I cannot apply it. In fact I see that the derivative $F[f]'(\lambda)$ can be seen, if we define $\varphi_n(x)=f(x)\frac{e^{-ih_n x}-1}{h_n}$, where $\{h_n\}$ converges to 0, as the limit $$\lim_n \int_{\mathbb{R}}f(x)\frac{e^{-i(\lambda+h_n)x}-e^{-i\lambda x}}{h_n}d\mu_x=\lim_n \int_{\mathbb{R}} f(x) \frac{e^{-ih_n x}-1}{h_n} e^{-i\lambda x} d\mu_x$$$$=\lim_n F[\varphi_n](\lambda)$$but I cannot prove that $\int_{\mathbb{R}} f(x) \frac{e^{-ih_n x}-1}{h_n} e^{-i\lambda x} d\mu_x$ converges to $-i\int_{\mathbb{R}}x f(x) e^{-i\lambda x} d\mu_x$. I thought about Lebesgue's dominated convergence theorem, but I am not convinced at all that we can majorate $|\varphi_n(x)|$ with a summable function and I am not sure this is the right path to prove the desired equality...  P.S.: I assume that the equality holds for any $f\in L_1(\mathbb{R},\mu_x)$ because the book does not specify any other conditions, but it is perfectly in the style of the book to state theorems valid only under certain conditions without explicitly declare them. So, if you noticed that some other condition is needed...
84143
Consider the series $ \sum_{n=1}^\infty x^2+ n/n^2$  . Pick out the true statements: (a) The series converges for all real values of $x$. (b) The series converges uniformly on $\mathbb{R}$. (c) The series does not converge absolutely for any real value of $x$.        stuck on this problem totally. please help me to solve this problem.thanks .
104399
I am trying to determine if $\sum_{i=1}^\infty \frac{n}{e^n}$ is converging  This is what I have so far  $\sum_{i=1}^\infty \frac{n}{e^n}$ converges by the geometric series test since $\frac{1}{e}$ &lt; 1   This seems too simple of answer and was wondering if this was even a geometric sequence.
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I need help with this exercise.  I need to prove $$\int_{0}^{1}x^{-x}\ dx=\sum_{n=1}^{\infty}n^{-n}$$  I think I should use some convergence theorem, but I'm stuck.  Thanks a lot!
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Per the title, if $\sum_{1}^{\infty}(a_n)^3$ diverges, does this imply that $\sum_{1}^{\infty}(a_n)$ diverges?   I'd appreciate hints (!) for dealing with this excercise.  EDIT Per the contrapositive, it is not given that $a_n$ converges absolutely, or that it is nonnegative for all $n$.  Thank you!
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I just need to know if this integral converges.  If it does, what is the value?  If not can anyone tell me what is the value of the $$ \int_{-\infty}^\infty 2\,\frac{\cos(w)\sin(w)}{w}\ dw ? $$
4075
I have to prove the  by tomorrow and I am really unconfident about what I did:  $$\sum_{n=1}^\infty a_n\text{ converges } \iff \sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}$$  I did the following:  Let $(b_n)$ be a sequence as follow: $b_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $a_{n+1}\leq a_n$ and so $0\leq a_{n+p}\leq a_n$ for all $n,p\in\mathbb N$.  So $\sum\limits_{n=1}^\infty b_n$ converges by the majorizing series $\sum\limits_{n=1}^\infty a_n$. And it's $\sum\limits_{n=0}^\infty b_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n+1}}=\sum\limits_{n=1}^\infty 2^{n-1}a_{2^n}$ so $\Rightarrow$ is done.  For $\Leftarrow$ consider $c_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $|a_n|\leq c_n$ and $\sum\limits_{n=0}^\infty c_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n}}=\sum\limits_{n=1}^\infty 2^{n}a_{2^n}$ and so $\sum\limits_{n=1}^\infty a_n$ converges by the majorizing series $\sum\limits_{n=0}^\infty c_n$.  Is this in form and content correct?
42656
Let $\{a_{mn}\},~m,n \in \Bbb{N}$, be an arbitrary double sequence of real numbers then does the following equality occur always?   $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{mn}^3=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}a_{mn}^3$$   I think there is no reason for the equality of this iterated series in general but I cannot figure out how to construct a counter example. Can anyone please help me how to create a example of $\{a_{mn}\}$ such that the given equality does not hold...!!
27650
It's a question from a test that I had, and I don't know how to prove this, so I am forwarding this to you.    $\sum \limits_{n=0}^{\infty }\:b_n$ is absolutely convergent series . How to prove that the series $\sum \limits_{n=0}^{\infty }\:(e^{b_n}\:-\:1)$ is also absolutely convergent?  We cannot assume here that $b_n\:\ge 0$ for every $n$.
52945
Consider the series of function $$\sum_{n=1}^{\infty}\frac{x}{1+n^2x}.$$  Show that this series of function is NOT uniformly convergent in $[0,1]$.   I know only two methods to show a series of function to NOT uniformly convergent :  (i) Using definition of uniform convergent.  (ii) Showing that term by term integration is not possible.  i.e. showing that$$\int_0^1\sum_{n=1}^{\infty}f_n(x)\,dx\not =\sum_{n=1}^{\infty}\int_0^1f_n(x)\,dx.$$  But in each case we need to find out the finite sum $\sum_{k=1}^{n}f_k(x)$.  I am not able to find out this finite sum.  Please help on it....  Does there any other process or theorem to find that the series is not uniformly convergent ?
45963
1) Construct a continuous function $f$ on $\mathbb{R}$ that is integrable on $\mathbb{R}$ but $\displaystyle\limsup_{x \to \infty} f(x)  = \infty$.  I took the function that is equal to $n$ on $[n, n + 1/n^{3})$ and made it continuous by saying that $f$ is the line segment joining $n$ and $n+1$ on $[n + 1/n^{3}, n+1)$. But I am failing to prove this integrable. For this, $\lim f(x) = \infty$ but how do you prove in general, if limit does not exist that $\limsup$ is infinity?  2) Prove that if $f$ is uniformly continuous and integrable on $\mathbb{R}$ we have $\displaystyle\lim_{|x| \to \infty} f(x) = 0$.  Any help is appreciated.  Thanks
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!
17450
I have difficulty to demonstrate this:  $$ \int_{-\infty}^{\infty}|f(x)|^2dx&lt;\infty~~~\text{(summable square function)}$$  then,   $$\lim_{|x|\rightarrow\infty }f(x)=0$$  thank you.
22436
Let $\sum_{n=1}^{\infty}a^2_n$ and $\sum_{n=1}^{\infty}b^2_n$ prove that: $\sum_{n=1}^{\infty}a_nb_n$ converges.  I thought about this answer: In order for a sum to converge i has to be bounded, Means Let $M_1\epsilon R$, So that $ \sum_{n=1}^{\infty}a_n&lt;M_1 $. Same goes for $b_n$. [And lets say $M_2$].  So is it just: $\sum_{n=1}^{\infty}a_nb_n\le \max(M_1*M_2, \sqrt{M_1*M_2}$)?  And one more question: Let $\sum_{n=1}^{\infty}a^2_n$ prove that  $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges.  If i go the same as before, I have hard time cause i get something like:  $\sum_{n=1}^{\infty}\frac{M_1 }{n}$ which doesn't converge [I think].  Any guidelines please?
10579
Consider the following problem from Axler's &quot;Measure, integration and real analysis&quot; (p72):  Honestly, I'm not even sure where to begin. This problem appears in the section that proves Lusin's and Egorov's theorem. The abstract Lebesgue integral is not yet introduced so I can not use that. Otherwise I could just show that the sum has finite integral and could conclude. I think we can write $f= \lim_k f_k$ as a pointwise limit of the partial sums but this probably does not quite help. Further attempt: I tried to show $$\{x: f(x) &lt; 1\}$$ contains a set of infinite measure but was unsuccesful. A hint to get started is appreciated!