# Limit of Lebesgue integrals in $L_1(\mathbb{R},m)$

Let $g\in L_1(\mathbb{R},m)$ bounded function where $m$ is the Lebesgue measure in $\mathbb{R}$. If

$$\lim_{x\to\pm\infty} g(x)=0,$$

show that for all functions $f\in L_1(\mathbb{R},m)$ we have:

$$\lim_{n\to\infty}\frac{1}{n}\int_{\mathbb{R}}g(x)\cdot f\left(\frac{x}{n}\right)\,dm=0.$$

Attempt: I tried exactly $2$ hours, I followed distinct ways but I have not idea how to do, I don't know how to apply Fatou's Lemma or LMCT or LDCT if is possible.

Let $x=ny$. Then $$\frac{1}{n}\int_{\mathbb{R}}g(x)f(\frac{x}{n})dx=\int_{\mathbb{R}}g(ny)f(y)dy.\tag{1}$$ Let $n\to\infty$ and apply dominated convergence theorem to the right hand of side $(1)$. The conclusion follows.
• +1. Note that since $g$ is bounded (say by $B$), $g(ny)f(y)$ is dominated by $Bf(y)$. May 9 '13 at 4:07
• I didn't get it, What means $dx$, must be $dm$? How to change of variable in Lebesgue integral? May 9 '13 at 4:08
• @GastónBurrull Change of variable works the same way for Lebesgue integral as for Riemann integral. $d x, dy, dm$ are all the same, the first two are used to indicate which letter is the variable (it is assumed in these cases the Lebesgue measure is being used). May 9 '13 at 4:30
• @GastónBurrull: Yes, the general discussion of change of variables is based on absolute continuity and fundamental theorem of calculus for Lebesgue integral. However, in this special case $\int_{\mathbb{R}}h(x)dx=n\int_{\mathbb{R}}h(nx)dx$, you may use continuous functions to $L^1$ approximate $h$, and apply change of variables to continuous functions.