Show that $f_n(t) = n\int_{t-1/n}^t1_{\{|f(s\vee 0)|\le n\}}f(s\vee 0)\,ds \rightarrow f(t)$

Question

Let $f: \mathbb R^+ \rightarrow \mathbb R$ be a left continuous function.

Show that $f_n(t) = n\int_{t-1/n}^t1_{\{|f(s\vee 0)|\le n\}}f(s\vee 0)\,ds \rightarrow f(t)$ for all $t\ge 0$, where $a\vee b= max\{a,b\}$

I really have no clue where to start.

$f_n(t)$ is the mean of the function $1_{\{|f(s\vee 0)|\le n\}}f(s\vee 0)$ on the interval $[t- 1/n, t]$.

Answer 1

Note that the statement is true for $t=0$. Therefore, we can assume that $t>0$. Since $f$ is left continious, for any $\varepsilon >0$ there exists an $\delta>0$, w.l.o.g. $0< t-\delta$, such that $|f(s) - f(t)|< \varepsilon$ for all $t-\delta \leq s \leq t$.

This implies that $f(s)$ is bounded on $[t-\delta,t]$. Taking $N \in \mathbb{N}$ so large that $1/N < \delta$ and $|f(s)| \leq N$ on $[t-\delta,t]$, we find for all $n \geq N$ $$|f_n(t)-f(t)| = | n \int_{t-1/n}^t (f(s)-f(t)) \mathop{dt}| \leq \max_{s \in [t-\delta,\delta]}|f(s)-f(t)| \leq \varepsilon.$$