How to switch the order of limit and integration here? Let $f\in L^1(\mathbb{R}^n)$ and $\phi_\varepsilon(x):=\varepsilon^{-n}e^{-\pi|\varepsilon^{-1}x|^2}$. It is straightforward to see that $\phi_\varepsilon(x)$ is an approximate identity.
I want to prove that 
$$\int_{\mathbb{R}^n}\left\vert\lim_{\varepsilon\to 0}\phi_\varepsilon*f-f\right\vert dx= \lim_{\varepsilon\to 0}\int_{\mathbb{R}^n}\left\vert\phi_\varepsilon*f-f\right\vert dx .$$
Can we apply Lebesgue Dominated Convergence theorem here? If not, could we find another way out, namely how to switch the order of limit and integration of
$$\lim_{\varepsilon\to 0}\int_{\mathbb{R}^n}\left\vert\phi_\varepsilon*f-f\right\vert dx?$$
If we apply Lebesgue Dominated Convergence theorem, then we need to do two things:


*

*Prove that $\phi_\varepsilon*f(x)-f(x)$ converges pointwise to a function $h(x)$ as $\varepsilon\to 0$.

*Prove that $\phi_\varepsilon*f(x)-f(x)$ is dominated by some integrable function $g(x)$ in the sense that
$$|\phi_\varepsilon*f(x)-f(x)|\le |g(x)|.$$
But I’m totally stuck here. Could you give me some help? Thank you!
Motivation:
I’m reading the proof of Fourier Inversion theorem:

$$\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi=f(x)$$
  almost everywhere, where $f,\widehat{f}\in L^1(\mathbb{R}^n)$.

I managed to understand the proof (including the equality right below) until
$$\lim_{\varepsilon\to 0}\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}e^{-\pi|\varepsilon \xi|^2}d\xi= \lim_{\varepsilon\to 0}\phi_\varepsilon*f(x).$$
It is mentioned in many books but there are always no more details. We are always told to apply Lebesgue Dominated Convergence theorem here and then the proof shall be concluded.
I managed to understand that the left side is equal to 
$$\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi.$$ For the left side, the theorem is indeed applied.
For the right side, if the equality that I stated on the top holds, then since
$$\lim_{\varepsilon\to 0}\int_{\mathbb{R}^n}\left\vert\phi_\varepsilon*f-f\right\vert dx=\lim_{\varepsilon\to 0}\Vert f*\phi_\varepsilon-f\Vert_1=0,$$we have $$\int_{\mathbb{R}^n}\left\vert\lim_{\varepsilon\to 0}\phi_\varepsilon*f-f\right\vert dx=0.$$
There is a theorem stating that

$E$ is a measurable set and $f$ is integrable on $E$. Then $$\int_E |f|~dm=0$$if and only if $f=0$ almost everywhere.

Therefore we have $\lim\limits_{\varepsilon\to 0}\phi_\varepsilon*f=f$ almost everywhere. It completes the proof.
Any help is appreciated. Thank you!
 A: The comments have that it is enough to show $\lim_{\epsilon \rightarrow 0} f*\phi_{\epsilon}(x) = f$ a.e., I would just like to show a solution that doesn't require Lebesgue Differentiation Theorem. First, the integral of $\phi_{\epsilon}$ over $\mathbb{R}^n$ is equal to 1 since
\begin{align*}
     \int_{\mathbb{R}^n} \phi_{\epsilon}(x) \ dx &= \epsilon^{-n}\int_{\mathbb{R}^n}e^{-\pi\epsilon^{-2} |x|^2}\ dx \\
&= \int_{\mathbb{R}^n}e^{-\pi|y|^2}dy\\
&= 1,
\end{align*}
where we employed the change of variable 
$$\epsilon y = x \implies \epsilon^n dy = dx,$$
and I also had to look at the wiki for the integral of an n-dimensional Gaussian.
EDIT: What is written below is not sufficient to prove this this theorem.
Now,
\begin{align*}
     \lim_{\epsilon \rightarrow 0} \left| f*\phi_{\epsilon}(x) - f(x)\right| &= \lim_{\epsilon \rightarrow 0} \left| \int_{\mathbb{R}^n} f(x-y)\phi_{\epsilon}(y) \ dy - f(x)\right|\\
&= \lim_{\epsilon \rightarrow 0} \left| \int_{\mathbb{R}^n} f(x-y)\phi_{\epsilon}(y) \ dy - \int_{\mathbb{R}^n} f(x)\phi_{\epsilon}(y) \ dy\right|\\
&\leq \lim_{\epsilon \rightarrow 0} \int_{\mathbb{R}^n} |\phi_{\epsilon}(y)|| f(x-y)- f(x)|\ dy.
\end{align*}
 Now, $\phi$ is clearly bounded and $f \in L^1(\mathbb{R}^n)$ so we can apply dominated convergence to pass the limit through the integral to get the desired result.
A: Let $B_r$ denote the ball of radius $r$ centered at the origin. Then Lebesgue Differentiation Theorem tells:

Theorem. Let $f$ be a locally integrable function on $\mathbb{R}^n$. Then
  $$\lim_{r \to 0} \frac{1}{|B_r|} \int_{B_r(x)} f(y) \, \mathrm{d}y = f(x) \tag{*}$$
  holds for Lebesgue-a.e. $x\in\mathbb{R}^n$.

Now let $f : \mathbb{R}^n \to \mathbb{R}$ be a measurable function which is either non-negative or integrable, and let $\phi_{\varepsilon}$ be as in OP. Then
\begin{align*}
(f \ast \phi_{\varepsilon})(x)
&= \int_{\mathbb{R}^n} f(x-y)\varepsilon^{-n}e^{-\pi|y/\varepsilon|^2} \, \mathrm{d}y \\
&= \int_{\mathbb{R}^n} f(x-y) \biggl( \int_{|y|}^{\infty} 2\pi r \varepsilon^{-n-2}e^{-\pi r^2 / \varepsilon^2} \, \mathrm{d}r \biggr) \, \mathrm{d}y \\
&= \int_{0}^{\infty} \biggl( \int_{\{|y| < r\}} f(x-y)  \, \mathrm{d}y \biggr) 2\pi r \varepsilon^{-n-2}e^{-\pi r^2 / \varepsilon^2} \, \mathrm{d}r \tag{$\because$ Fubini} \\
&= \int_{0}^{\infty} \biggl( \frac{1}{|B_{\varepsilon s}|} \int_{B_{\varepsilon s}(x)} f(z)  \, \mathrm{d}z \biggr) 2\pi |B_1| s^{n+1} e^{-\pi s^2} \, \mathrm{d}s,
\end{align*}
where the substitutions $r=\varepsilon s$ and $z=x-y$ are utilized in the final step. Write
$$k(s) := 2\pi |B_1| s^{n+1} e^{-\pi s^2}$$
for simplicity. Then either by a direct computation or by plugging $f \equiv 1$ to the identity above, we note that
$$\int_{0}^{\infty} k(s) \, \mathrm{d}s = 1.$$
Now assume that $f$ is integrable. Then for each $x\in\mathbb{R}^n$ satisfying $\text{(*)}$, the map
$$ (0, \infty) \ni r \mapsto \frac{1}{|B_{r}|} \int_{B_{r}(x)} f(z)  \, \mathrm{d}z $$
is bounded and converges to $f(x)$ as $r \to 0$. So by the dominated convergence theorem,
\begin{align*}
\lim_{\varepsilon \to 0} (f \ast \phi_{\varepsilon})(x)
&= \int_{0}^{\infty} \biggl( \lim_{\varepsilon \to 0} \frac{1}{|B_{\varepsilon s}|} \int_{B_{\varepsilon s}(x)} f(z)  \, \mathrm{d}z \biggr) k(s) \, \mathrm{d}s \\
&= \int_{0}^{\infty} f(x)k(s) \, \mathrm{d}s
= f(x)
\end{align*}
as required.
