# On Constructing Functions, part 1

/- $f_n$ converges uniformly
- $f_n$ converges pointwise
- $f_n$ converges almost everywhere (a.e.)
- $f_n$ converges in $L^1$ (set of Lebesgue integrable functions)
- and so on...

*not*have the memory of an elephant (whatever that phrase means...), I've decided to keep a list of different sequences that converge (or don't converge) to different functions in different ways. With each example I'll also include a little (and hopefully) intuitive explanation for

*why*. Having these sequences close at hand is especially useful when analyzing the behavior of certain functions or constructing counterexamples.

The first sequence we'll look at is one which converges almost everywhere, but does not converge in $L^1$ (the set of Lebesgue integrable functions).

*Also in this series*:

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

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

Example 4: $f_n$ are integrable and converge uniformly to $f$, yet $f$ is not integrable

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

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

## Example 1

**A sequence of functions $\{f_n:\mathbb{R}\to\mathbb{R}\}$ which converges to 0 almost everywhere but does not converge to 0 in $L^1$.**

**This works because:** Recall that to say $f_n\to 0$ almost everywhere means $f_n\to 0$ * pointwise on $\mathbb{R}$ except for a set of measure 0.* Here, the set of measure zero is the singleton set $\{0\}$ (at $x=0$, $f_n(x)=n$ and we can't make this less than $\epsilon$ for *any* $\epsilon >0$). So $f_n$ converges to 0 pointwise on $(0,1]$. This holds because if $x< 0$ or $x>1$ then $f_n(x)=0$ for all $n$. Otherwise, if $x\in(0,1]$, we can choose $n$ appropriately:

**The details:** Let $\epsilon>0$ and $x\in (0,1]$ and choose $N\in \mathbb{N}$ so that $N>\frac{1}{x}$. Then whenever $n>N$, we have $n>\frac{1}{x}$ which implies $x>\frac{1}{n}$ and so $f_n(x)=0$. Hence $|f_n{x}-0|=0< \epsilon$.

Further*, $f_n\not\to 0$ in $L^1$ since $$\int_{\mathbb{R}}|f_n|=\int_{[0,\frac{1}{n}]}n=n\lambda([0,\frac{1}{n}])=1.$$

**Remark:**Notice that Egoroff's theorem applies here! We just proved that $f_n\to 0$ pointwise a.e. on $\mathbb{R}$, but Egoroff says that we can actually get uniform convergence a.e. on a bounded subset of $\mathbb{R}$, say $(0,1]$.

In particular for each $\epsilon >0$ we are guaranteed the existence of a subset $E\subset (0,1]$ such that $f_n\to 0$ uniformly and $\lambda((0,1]\smallsetminus E)<\epsilon$. In fact, it should be clear that that subset must be something like $(\frac{\epsilon}{2},1]$ (the "zero region" in the graph above). Then no matter where $x$ is in $(0,1]$, we can always find $n$ large enough - namely all $n$ which satisfy $\frac{1}{n}<\frac{\epsilon}{2}$ - so that $f_n(x)=0$, i.e. $f_n\to f$ uniformly. And indeed, $\lambda((0,1]\smallsetminus (\frac{\epsilon}{2},1]=\epsilon/2<\epsilon$ as claimed.

For a measurable set $X\subset \mathbb{R}$, denote the set of all Lebesgue integrable functions $f:X\to\mathbb{R}$ by $L^1(X)$. Then a sequence of functions $\{f_n\}$ is said to

*converge in $L^1$*to a function $f$ if $\displaystyle{\lim_{n\to\infty}}\int|f_n-f|=0$.