# Pointwise limit of integrable function

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.

• You should be a bit more specific by what you mean by integrable. – xavierm02 Mar 15 '13 at 12:35
• To answer your question, yes. Check out dominated convergence theorem for details. – Gautam Shenoy Mar 15 '13 at 13:17

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 $\mathbb{Z}$.

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).

• Aguirre: Might be a stupid question. Can you please explain the last equality. Thanks – Mr. MBB Jul 29 '16 at 1:50
• 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 $1$. – Julián Aguirre Jul 29 '16 at 8:36

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 ...

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.