Intro to Probability - YAQ Week 6

Question

We know that $P(\emptyset)=0$ for any probability $P$. Can we say that $E=\emptyset$ if $P(E)=0$?

While the answer is sometimes yes, like for example in a finite sample space where all outcomes have positive probability (like in the case equiprobable outcomes), the answer is no in general. To see this, consider flipping a fair coin repeatedly. Consider the event $\omega$ of the coin always landing on $H$ each time, i.e $\omega=(H,H,H,H,\dots)$. We show that $P(\omega)=P(\{ \omega\})=0$, while, of course, $\{ \omega \}\neq\emptyset$. Notice that, instead of $\omega$, we could have chosen any specific sequence of heads and tails. Define the events $$ E_n=\big\{ (F_1,F_2,\dots)\in \{ H,T \}^\mathbb{N}\, \big |\, F_1=F_2=\dots=F_n=H\big \} $$ for $n\in \mathbb{N}$ and observe that $P(E_n)=\frac{1}{2^n}$. It then follows that $$ P(\{ \omega \})=P \bigl( \bigcap _{n=1}^\infty E_n\bigr)= \lim _{n\to\infty}P(E_n)=\lim_{n\to\infty}\frac{1}{2^n}=0 $$ since $E_{n+1}\subset E_n$ for $n\geq 1$.

Question

Why does it hold that $\frac{1}{k}\sum_{i=0}^k\bigl(\frac{i}{k}\bigr)^m\simeq \frac{1}{m+1}$?

The integral of a continuous function $f:[0,1]\to \mathbb{R}$ can be computed as a limit of Riemann sums. In particular, if you partition the interval $[0,1]$ into subintervals $$ J_i=\bigl[\frac{i-1}{k},\frac{i}{k}\bigr),\: i=1,\dots,k $$ and choose $x_i\in J_i$, then $$ \int _0^1 f(x)\ dx=\lim _{k\to\infty} \sum _{i=1}^k f(x_i)\frac{1}{k}, $$ where, of course, $ \frac{1}{k}=\frac{i}{k}-\frac{i-1}{k}$ is the length of $J_i$. Now simply choose $f(x)=x^m$ to see that $$ \frac{1}{m+1}=\frac{1}{m+1}x^{m+1}\Big |_0^1=\int _0^1 x^m\ dx= \lim _{k\to\infty} \sum _{i=1}^k x_i^m\frac{1}{k}= \lim _{k\to\infty} \sum _{i=1}^k \bigl( \frac{i}{k}\bigr) ^m\frac{1}{k}= \lim _{k\to\infty} \frac{1}{k}\sum _{i=0}^k \bigl( \frac{i}{k}\bigr) ^m, $$ where we used $x_i=\frac{i}{k}$, $i=0,\dots, k$.