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

my approach in brief: since $f,g\in {L}^{1}$, by Tonelli-Fubini's theorem, we can show

${\int}_{\mathbb{R}}{f}_{t}\ast gdx=({\int}_{\mathbb{R}}f(x)dx){\int}_{\mathbb{R}}g(y)dy$

${\int}_{\mathbb{R}}{f}_{t}\ast gdx=({\int}_{\mathbb{R}}f(x)dx){\int}_{\mathbb{R}}g(x)dx$

${\int}_{\mathbb{R}}({f}_{t}\ast g-({\int}_{\mathbb{R}}f(x)dx){\int}_{\mathbb{R}}g(x))dx=0$

Therefore,

${\int}_{\mathbb{R}}|{f}_{t}\ast g(x)-ag(x)|dx=0$

I am feeling something is wrong with my approach. Do correct me and give me some hints for solving this simple question.

my approach in brief: since $f,g\in {L}^{1}$, by Tonelli-Fubini's theorem, we can show

${\int}_{\mathbb{R}}{f}_{t}\ast gdx=({\int}_{\mathbb{R}}f(x)dx){\int}_{\mathbb{R}}g(y)dy$

${\int}_{\mathbb{R}}{f}_{t}\ast gdx=({\int}_{\mathbb{R}}f(x)dx){\int}_{\mathbb{R}}g(x)dx$

${\int}_{\mathbb{R}}({f}_{t}\ast g-({\int}_{\mathbb{R}}f(x)dx){\int}_{\mathbb{R}}g(x))dx=0$

Therefore,

${\int}_{\mathbb{R}}|{f}_{t}\ast g(x)-ag(x)|dx=0$

I am feeling something is wrong with my approach. Do correct me and give me some hints for solving this simple question.