Proof that the Cesaro mean converges using integration with respect to counting measure I have seen a lot of proofs of the following:
Suppose $a_n \to a$, then $\frac{a_1 + \ldots + a_n}{n} \to a$. 
However, it seems to me that there could be a proof along these lines.
Let $\nu$ be counting measure on $\mathbb{N}$. Let $$f_n : \mathbb{N} \to \mathbb{R} \quad \text{   be   } \quad f_n (i) = \frac{a_i}{n} \chi_{[0,n]}$$ Then calculate the integral of $f_n$ with respect to $\nu$ and take the limit as $n \to \infty$. I am not sure how to do this last step. I am thinking that maybe I should use the Monotone Convergence Theorem or the Dominated Convergence Theorem.
 A: Unfortunatelly, it cannot be proven this way.
To see this, define $f_n : \mathbb{N} \to \mathbb{R}$ like $f_n(m) = \frac{a_m}{n}\chi_{[0,n]}(m)$, $\forall m \in \mathbb{N}$ and suppose that you can switch the limit and the integral (btw. check your definition of $f_n$, perhaps you made a typo). 
\begin{align}
a &= \lim_{n\to\infty} \frac{1}{n}\sum_{m=1}^n a_m \\ &= \lim_{n\to\infty}\sum_{m=1}^\infty \frac{a_m}{n}\chi_{[0,n]} \\ 
&=\lim_{n\to\infty}\sum_{m=1}^\infty f_n(m) \\
&= \lim_{n\to\infty}\int_\mathbb{N} f_n \,d\nu \\
&= \int_\mathbb{N} \lim_{n\to\infty} f_n \,d\nu \\
&= \int_\mathbb{N} 0 \,d\nu \\
&= 0
\end{align}
We arrive at a contradiction.
The assumption we are missing is $\sum_{m=1}^\infty \frac{|a_m|}{m} < +\infty$:
$$|f_n(m)| = \frac{|a_m|}{n}\chi_{[0, n]} \leq \frac{|a_m|}{m}$$
since $m \leq n$ if $f_n(m) \ne 0$.
In general $\lim_{n\to\infty} {a_n} = a$ does not imply $\sum_{m=1}^\infty \frac{|a_m|}{m} < +\infty$ (e.g. take take $a_n = 1$, $\forall n\in\mathbb{N}$), so we cannot dominate the functions $f_n$ in general.
