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let $f$,$g$ be ${L}^{1}\left(\mathbb{R}\right)$ functions with Lebesgue measure. Define ${f}_{t}\left(x\right)=\frac{f\left(x/t\right)}{t}$. Prove ${f}_{t}\ast g$ converges to $ag$ in ${L}^{1}$ when $t\to {0}^{+}$, where $a={\int }_{\mathbb{R}}f\left(x\right)dx$.

my approach in brief: since $f,g\in {L}^{1}$, by Tonelli-Fubini's theorem, we can show
${\int }_{\mathbb{R}}{f}_{t}\ast gdx=\left({\int }_{\mathbb{R}}f\left(x\right)dx\right){\int }_{\mathbb{R}}g\left(y\right)dy$
${\int }_{\mathbb{R}}{f}_{t}\ast gdx=\left({\int }_{\mathbb{R}}f\left(x\right)dx\right){\int }_{\mathbb{R}}g\left(x\right)dx$
${\int }_{\mathbb{R}}\left({f}_{t}\ast g-\left({\int }_{\mathbb{R}}f\left(x\right)dx\right){\int }_{\mathbb{R}}g\left(x\right)\right)dx=0$
Therefore,
${\int }_{\mathbb{R}}|{f}_{t}\ast g\left(x\right)-ag\left(x\right)|dx=0$
I am feeling something is wrong with my approach. Do correct me and give me some hints for solving this simple question.
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Paxton James
We have
$q\left(t\right):={\int }_{\mathbb{R}}^{}|{f}_{t}\ast g\left(x\right)-ag\left(x\right)|dx⩽{\int }_{\mathbb{R}}^{}|{f}_{t}\left(y\right)|\left({\int }_{\mathbb{R}}^{}|g\left(x-y\right)-g\left(x\right)|dx\right)dy:={\int }_{\mathbb{R}}^{}|{f}_{t}\left(y\right)|\phi \left(y\right)dy$
Suppose that $g$ is the indicator function of an interval $\left[\alpha ,\beta \right]$, then , hence
$q\left(t\right)⩽2{\int }_{\mathbb{R}}|f\left(s\right)|\text{min}\left(t|s|,\beta -\alpha \right)ds$
which converges to 0 by the monotone convergence theorem when $t\to {0}^{+}$.
The general result follows by density of the space of step functions in ${L}^{1}\left(\mathbb{R}\right)$.
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April Bush
The last inequality is a clever step.