# Lebesgue Integration

Can you help me with the following question?

Suppose $g \in L^1 ([0,\infty))$ and $\displaystyle\int_0^\infty g(x)dx =1$. Prove that if $f:[0,\infty)\rightarrow \mathbb{R}$ is a continuous function then $n \displaystyle\int_0^1 f(x+t) g(nt) dt \rightarrow f(x)$ as $n\rightarrow \infty$ for $x\in \mathbb{R}$.

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## 1 Answer

For fixed $x\in\mathbb{R}$, let $y=nt$, then $$\tag{1}n \displaystyle\int_0^1 f(x+t) g(nt) dt=\int_0^nf(x+\frac{y}{n})g(y)dy=\int_0^\infty \chi_{[0,n)}(y)f(x+\frac{y}{n})g(y)dy.$$ Here $\chi_{[0,n)}$ is the characteristic function: $$\chi_{[0,n)}(y)= \begin{cases} 1, & y\in[0,n)\hbox{;} \\ 0, & \hbox{otherwise.} \end{cases}$$

For $y\in[0,n)$, $x\leq x+\displaystyle\frac{y}{n}< x+1$. Since $f$ is continuous on $[x,x+1]$, let $M=\max_{z\in[x,x+1]}|f(z)|<\infty$, then $$\big|\chi_{[0,n)}f(x+\frac{y}{n})g(y)\big|\leq M|g(y)|\in L^1([0,\infty)).$$ Therefore, by Dominated convergence theorem, we have by $(1)$ $$\lim_{n\rightarrow\infty}n \displaystyle\int_0^1 f(x+t) g(nt) dt=\int_0^\infty \lim_{n\rightarrow\infty}\chi_{[0,n)}(y)f(x+\frac{y}{n})g(y)dy$$ $$= \int_0^\infty f(x)g(y)dy=f(x)\int_0^\infty g(y)dy=f(x)$$ Here we have used the assumption that $f$ is continuous, which implies that $\lim_{n\rightarrow\infty}f(x+\frac{y}{n})=f(x)$ for all $y\in[0,n)$.

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