# On Constructing Functions, Part 4

This post is the fourth example in an ongoing list of various sequences of functions which converge to different things in different ways.

*Also in this series:*

Example 1: converges almost everywhere but not in $L^1$

Example 2: converges uniformly but not in $L^1$

Example 3: converges in $L^1$ but not uniformly

Example 5: converges pointwise but not in $L^1$

Example 6: converges in $L^1$ but does not converge anywhere

## Example 4

**A sequence of (Lebesgue) integrable functions $f_n:\mathbb{R}\to[0,\infty)$ so that $\{f_n\}$ converges to $f:\mathbb{R}\to[0,\infty)$ uniformly, yet $f$ is not (Lebesgue) integrable. **

Our first observation is that "$f$ is not (Lebesgue) integrable" can mean one of two things: either $f$ is not measurable or $\int f=\infty$. The latter tends to be easier to think about, so we'll do just that. Now *what function do you know of such that when you "sum it up" you get infinity?* How about something that behaves like the divergent geometric series? Say, its continuous cousin $f(x)=\frac{1}{x}$? That should work since we know $$\int_{\mathbb{R}}\frac{1}{x}=\int_{1}^\infty \frac{1}{x}=\infty.$$
Now we need to construct a sequence of integrable functions $\{f_n\}$ whose uniform limit is $\frac{1}{x}$. Let's think simple: think of drawring the graph of $f(x)$ one "integral piece" at a time. In other words, define:

**This works because:** It makes sense to define the $f_n$ as $f(x)=\frac{1}{x}$ "chunk by chunk" since this way the convergence is guaranteed to be uniform. Why? Because how far out we need to go in the sequence so that the difference $f(x)-f_n(x)$ is less than $\epsilon$ *only* depends on how small (or large) $\epsilon$ is. The location of $x$ doesn't matter!

Also notice we have to define $f_n(x)=0$ for all $x< 1$ to avoid the trouble spot $\ln(0)$ in the integral $\int f_n$. This also ensures that the area under each $f_n$ is finite, guaranteeing integrability.

**The details:** Each $f_n$ is integrable since for a fixed $n$,
$$\int_{\mathbb{R}}f_n=\int_1^n\frac{1}{x}=\ln(n).$$
To see $f_n\to f$ uniformly, let $\epsilon >0$ and choose $N$ so that $N>1/\epsilon$. Let $x\in \mathbb{R}$. If $x\leq 1$, any $n$ will do, so suppose $x>1$ and let $n>N$. If $1< x \leq n$, then we have $|f_n(x)-f(x)|=0< \epsilon$. And if $x> n$, then $$\big|\frac{1}{x}\chi_{[1,\infty)}(x)-\frac{1}{x}\chi_{[1,n]}(x)\big|=\big|\frac{1}{x}-0\big|=\frac{1}{x}<\frac{1}{n}< \frac{1}{N}< \epsilon.$$

Voila!