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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.

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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.

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  • $\begingroup$ +1. Note that since $g$ is bounded (say by $B$), $g(ny)f(y)$ is dominated by $Bf(y)$. $\endgroup$ May 9 '13 at 4:07
  • $\begingroup$ I didn't get it, What means $dx$, must be $dm$? How to change of variable in Lebesgue integral? $\endgroup$ May 9 '13 at 4:08
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    $\begingroup$ @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). $\endgroup$ May 9 '13 at 4:30
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    $\begingroup$ @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. $\endgroup$
    – 23rd
    May 9 '13 at 5:17
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    $\begingroup$ @AlexBecker: Thank you for helping me to answer the OP's questions. I answered his/her last question. $\endgroup$
    – 23rd
    May 9 '13 at 5:19

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