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 $$\big|\frac{1}{x}\chi_{[1,\infty)}(x)-\frac{1}{x}\chi_{[1,n]}(x)\big|=\big|\frac{1}{x}-0\big|=\frac{1}{x}<\frac{1}{n}< \frac{1}{N}< \epsilon.$$