Need a help in an example of a bounded linear operator with estimates of the norm. The example is given in the following pictures:



I have a difficulty in understanding the last line, I know that a function is continuous at a point $t_{0}$ iff the following definition is satisfied:$$\begin{equation*}
\lim_{x \rightarrow t_{0}}
f(x) = f(t_{0})
\end{equation*}$$
But I did not understand the relation between this definition and the definition he used in the last line, how is the limit transformed into integration? could anyone clarify this for me please?  
 A: So $\varphi:t\rightarrow|a(t)|^{2}$ is a real-valued continuous function on $[a,b]$, by Extreme Value Theorem, $\varphi$ attains both minimum and maximum on $[t_{0}-1/n,t_{0}+1/n]$, say, $m=\min=\varphi(\alpha_{n})$, $M=\max=\varphi(\beta_{n})$, so 
\begin{align*}
\varphi(\alpha_{n})=\dfrac{n}{2}\int_{t_{0}-1/n}^{t_{0}+1/n}\varphi(\alpha_{n})dt\leq\dfrac{n}{2}\int_{t_{0}-1/n}^{t_{0}+1/n}\varphi(t)dt\leq\dfrac{n}{2}\int_{t_{0}-n/2}^{t_{0}+n/2}\varphi(\beta_{n})dt=\varphi(\beta_{n}),
\end{align*}
then by Intermediate Value Theorem we have some $\eta_{n}$ lies in between $\alpha_{n}$ and $\beta_{n}$ such that 
\begin{align*}
\varphi(\eta_{n})=\dfrac{n}{2}\int_{t_{0}-1/n}^{t_{0}+1/n}\varphi(t)dt,
\end{align*}
as $\eta_{n}\rightarrow t_{0}$ whenever $n\rightarrow\infty$ and $\varphi$ is continuous, so 
\begin{align*}
\dfrac{n}{2}\int_{t_{0}-1/n}^{t_{0}+1/n}\varphi(t)dt\rightarrow\varphi(t_{0}).
\end{align*}
A: Let $g(t)=|a(t)|^{2}$. We have to show that $\frac n 2 \int_{t_0 -1/n}^{t_0 +1/n} g(t)dt \to g(t_0)$. This is same as showing that  $\frac n 2 \int_{t_0 -1/n}^{t_0 +1/n} \{g(t)-g(t_0)\}dt \to 0$. Given $\epsilon >0$ choose n so large that $|g(t)-g(t_0)|< \epsilon$ whenever $|t-t_0|<1/n$. Then $|\frac n 2 \int_{t_0 -1/n}^{t_0 +1/n} \{g(t)-g(t_0)\}dt|\leq \epsilon$.
