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Question:

Show that the pointwise limit of integrable functions is not necessarily integrable.

I am stuck on this question. Here is what I know.

Let $(f_n)^{\infty}_{n=1}$ be a series of integrable functions, and let

$$\lim_{n \to \infty} f_n(x)=Z$$

I need to show that $Z$ is not necessarily integrable. Should I be looking for a specific example? A function that is integrable, but as $n \to \infty $ the function is no longer integrable.

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  • $\begingroup$ You should be a bit more specific by what you mean by integrable. $\endgroup$ – xavierm02 Mar 15 '13 at 12:35
  • $\begingroup$ To answer your question, yes. Check out dominated convergence theorem for details. $\endgroup$ – Gautam Shenoy Mar 15 '13 at 13:17
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Here is another natural example on $\mathbb{R}$ with the Lebesgue measure, which works also on $\mathbb{Z}$ with the counting measure: $$ f_n=1_{[-n,n]} $$ Use the Monotone Convergence Theorem.

In the same spirit, we can do even worse. Take $$ g_n=-1_{[-n,0)}+1_{(0,n]}. $$ Then all the $g_n$'s have $0$ integral. So $\lim _n \int g_n=0$ exists. Yet the pointwise limit is not integrable. Again neither on $\mathbb{R}$ nor on $\mathbb{Z}$.

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Just in case your question is about Riemann integrability, I will provide a sequence of Riemann integrable functions $f_n\colon[0,1]\to\mathbb{R}$ such that $\lim_{n\to\infty}f_n(x)=f(x)$ exists for all $x\in[0,1]$ but $f$ is not integrable in the sense of Riemann.

Order the rationals in $[0,1]$: $$ [0,1]\cap\mathbb{Q}=\{r_n\}_{n=1}^\infty. $$ Let $$ \phi_n(x)=\begin{cases} 1 &\text{if }x=r_n,\\ 0 &\text{otherwise} \end{cases} \quad\text{and}\quad f_n(x)=\sum_{k=1}^n\phi_k(x). $$ Then $f_n$ is Riemann integrable, since it has a finite number of discontinuities ($r_1,\dots,r_n$), but $$ f(x)=\lim_{n\to\infty}f_n(x)=\sum_{k=1}^\infty\phi_k(x)=\begin{cases} 1 &\text{if }x\text{ is rational}\\ 0 &\text{if }x\text{ is irrational} \end{cases} $$ is not Riemann integrable (althoug it is Lebesgue integrable).

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  • $\begingroup$ Aguirre: Might be a stupid question. Can you please explain the last equality. Thanks $\endgroup$ – Mr. MBB Jul 29 '16 at 1:50
  • $\begingroup$ If $x$ is irrational then $\phi_k(x)=0$ for all $k$ and the sum is $0$. If $x$ is rational then $x=r_n$ for some $n$, and all the terms in the sum are $0$ except the corresponding to $k=n$ which is equal to $1$. $\endgroup$ – Julián Aguirre Jul 29 '16 at 8:36
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Yes, a specific example is what you need, and you might start by looking at a well-known function which is not integrable (for example $f(x)=1/x$) and then find some integrable functions which converge to $f$ - e.g. consider the functions $f_n(x) = \max(f(x),n)$. I will let you check the details ...

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Check this,

$$ \int_{0}^{1}x^{-1+\frac{1}{n}}dx. $$

Note that, the functions $x^{1+1/n}$ are integrable as improper integrals or in the extended sense of Riemann and Lebesgue.

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