## Flashback to Undergrad Analysis (2012)

As I leisurely browsed through my old emails, I ran into a short sequence of them from a professor of a class I took in the early days of 2012. It was titled "Analysis 2", and I recall it being, without a shadow of a doubt, the most demanding one I took as an undergrad, by far. He would habitually send us polished proofs of claims we discussed in class, and as I browsed through them, although simple from my current standpoint, it escapes me how he expected an undergrad to produce them during an exam!

I picked two claims which I liked from his list and want to share them and their abridged proofs here.

####
Functions discontinuous at countably many points __can__ be Riemann integrable.

Let $f$ be such that $f\left(x\right)=1$ if $x=n^{-1}$ for some $n\in\mathbb N$ and $f\left(x\right)=0$ otherwise.

**Claim**: $f$ is Riemann integrable on $\left[0,1\right]$.**Proof**: Clearly, $L\left(f,P\right)=0$. Let $P_N=\left\{0,1/N,x_1,x_2,\dots,x_{N^2-N}\right\}$, where $x_k=1/N+k/N^2$. Then $U\left(f,P_N\right)=2/N$. Let $N>2/\varepsilon$. Then
$U\left(f,P_N\right)-L\left(f,P_N\right)=2/N<\varepsilon$. Moreover, as a bonus fact, it follows that $\int_0^1f=0$. $\blacksquare$

#### Moving limits into integrals is a delicate matter.

Let $g_n,g:\left[0,1\right]\to\mathbb R$ be integrable and bounded by $M$, and let $g_n\to g$ pointwise.

$$\begin{align}\left|\int_0^1g_n-\int_0^1g\right|&=\left|\int_0^1\left(g_n-g\right)\right|\\&=\left|\int_0^\delta\left(g_n-g\right)+\int_\delta^1\left(g_n-g\right)\right|\\&\leq\int_0^\delta\left|g_n-g\right|+\int_\delta^1\left|g_n-g\right|\\&<2M\delta+\frac\varepsilon2.\end{align}$$Letting $\delta=\varepsilon/4M$ yields the desired result. $\blacksquare$

**Claim**: If $g_n\to g$ uniformly on any $\left[\delta,1\right]$, where $\delta\in\left(0,1\right)$, then $\lim\int_0^1g_n=\int_0^1g.$**Proof**: Uniform convergence implies that $\left|g_n\left(x\right)-g\left(x\right)\right|<\varepsilon/2$ whenever $x\in\left[\delta,1\right]$ and $n\geq N$. Therefore,$$\begin{align}\left|\int_0^1g_n-\int_0^1g\right|&=\left|\int_0^1\left(g_n-g\right)\right|\\&=\left|\int_0^\delta\left(g_n-g\right)+\int_\delta^1\left(g_n-g\right)\right|\\&\leq\int_0^\delta\left|g_n-g\right|+\int_\delta^1\left|g_n-g\right|\\&<2M\delta+\frac\varepsilon2.\end{align}$$Letting $\delta=\varepsilon/4M$ yields the desired result. $\blacksquare$

Flashback to Undergrad Analysis (2012)
Reviewed by Josué Molina
on
2:49 AM
Rating: 5