How to prove that $f \in L^1_{loc} (\mathbb{R}^n)$ is constant a.e. if the regularizations $f_{\epsilon}$ are constant $\forall \epsilon$? Let $f \in L^1_{loc} (\mathbb{R}^n)$ and $f$ bounded. First I clarify the notation: Let $\rho \in C^{\infty}_c(\mathbb{R}^n)$ be a regularizing kernel, that is it satisfies
$$ spt \rho \subset \overline{B(0,1)},  \quad \rho(x) \geq 0 \quad \forall x \in \mathbb{R}^n, \quad \int_{\mathbb{R}^n} \rho(x) \,dx =1 \quad, \quad \rho (x) = \rho(-x) \quad \forall x \in \mathbb{R}^n.$$
and let $(\rho_{\epsilon})_{\epsilon\in (0,1)}$ be the corrispective family of mollifiers, that is $\forall \epsilon \in (0,1)$
$$\rho_{\epsilon}(x) := \frac{1}{\epsilon^n} \rho\left( \frac{x}{\epsilon}\right) \qquad \forall x \in \mathbb{R}^n.$$
Denote with $ f_{\epsilon} = f * \rho_{\epsilon}$ the convolution.
I have this problem. Suppose that $\forall \epsilon \in (0,1)$ exists a constant $C_\epsilon \in \mathbb{R}^n$ such that $f_{\epsilon} = C_\epsilon$. I have to prove that there exists a constant $C$ such that $f=C$ a.e. on $\mathbb{R}^n$, but I don't know how to prove that. This is my idea:
I think I should use the fact that a.e. $x \in\mathbb{R}^n $ is a Lebesgue point of $f$, so
$$ f(x) = \lim_{\epsilon \to 0^+} \frac{1}{|B(x, \epsilon)|}\int_{B(x, \epsilon)} f(y) \, dy. $$
If I could prove that
$$\lim_{\epsilon \to 0^+} \frac{1}{|B(x, \epsilon)|}\int_{B(x, \epsilon)} f(y) \, dy  \stackrel{(?)}{=} \lim_{\epsilon \to 0^+} \int_{\mathbb{R}^n} f(y) \rho_{\epsilon} (x-y) dy = \lim_{\epsilon \to 0^+} f_{\epsilon}(x) \quad \text{for a.e. }x $$
then I could conclude my proof by contraddicition.
Indeed I would have $f(x) = \lim_{\epsilon \to 0^+} C_{\epsilon}$ for a.e. $x$. BUT is the equality $\stackrel{(?)}{=}$ correct and how can I prove it?
Any advice is really appreciated! Thanks a lot!
 A: I don't think that it is necessary to use Lebesgue points at all. You probably know that $f_{\varepsilon} \to f $ in $\mathcal{D}^{\prime}(\mathbb{R}^n) \text{ as } \varepsilon \to 0\, ,$
that is, for all test functions $\varphi \in C_{c}^{\infty}(\mathbb{R}^n)$ the equation
\begin{equation}
\lim_{\varepsilon \to 0} \int_{\mathbb{R}^n} f_{\varepsilon}(x) \varphi(x) \, \text{d}x = \int_{\mathbb{R}^n} f(x) \varphi(x) \, \text{d} x\, \quad (1)
\end{equation}
holds. In particular, the limit on the left-hand side exists and is finite. As $f_{\varepsilon} = C_{\varepsilon}$, we can deduce that
$$
\infty > \lim_{\varepsilon \to 0} \int_{\mathbb{R}^n} f_{\varepsilon}(x) \varphi(x) \, \text{d} x = \big(\lim_{\varepsilon \to 0} C_{\varepsilon} \big)\int_{\mathbb{R}^n} \varphi(x) \, \text{d} x\, . \quad (2)
$$
Setting $C= \lim_{\varepsilon \to 0} C_{\varepsilon} $ and combing $(1)$ and $(2)$, we find that
$$
\int_{\mathbb{R}^n} \big(f(x) - C \big) \varphi(x) \, \text{d} x = 0 \quad \text{ for all } \varphi \in C_c^{\infty}(\mathbb{R}^n)\, .
$$
From this you can conclude that $f(x) = C$ for almost all $x \in \mathbb{R}^n$.
A: Suppose $f$ is not constant a.e.  Then there are two balls $A, B$ of the same radius $R > 0$ such that $\int_A f \ne \int_B f$ (e.g. these might be balls centred at Lebesgue points).  Say $\left| \int_A f - \int_B f\right| = \eta$.  Now approximate these integrals by integrals of $f_\epsilon$ and get a contradiction if $\epsilon$ is sufficiently small.
A: If you want to show the first equality you have, you can do it as follows: if $f_{x,\epsilon}=\frac{1}{|B(x,\epsilon)|}\int_{B(x,\epsilon)}f$, note that $$\begin{align*}\int_{\mathbb{R}^n}f(y)\rho_{\epsilon}(x-y)\,dy-f_{x,\epsilon}&=\int_{B(x,\epsilon)}f(y)\left(\rho_{\epsilon}(x-y)-\frac{1}{|B(x,\epsilon)|}\right)\,dy\\ &=\int_{B(x,\epsilon)}f(y)\left(\frac{1}{\epsilon^n}\rho\left(\frac{x-y}{\epsilon}\right)-\frac{1}{|B(x,\epsilon)|}\right)\,dy\\ &=\epsilon^{-n}\int_{B(x,\epsilon)}f(y)\left(\rho\left(\frac{x-y}{\epsilon}\right)-c\right)\,dy,\end{align*}$$ where $c=\frac{1}{|B(0,1)|}$. Setting $y=\epsilon z+x$, we obtain that $$\int_{\mathbb{R}^n}f(y)\rho_{\epsilon}(x-y)\,dy-\frac{1}{|B(x,\epsilon)|}\int_{B(x,\epsilon)}f(y)\,dy=\int_{B(0,1)}f(\epsilon z+x)(\rho(-z)-c)\,dz.$$ Letting $\epsilon\to 0$, the last expression goes to $$\int_{B(0,1)}f(x)(\rho(-z)-c)\,dz=f(x)\left(\int_{B(0,1)}\rho(-z)\,dz-c|B(0,1)|\right),$$ and the last expression is equal to $0$.
Edit: to show convergence as $\epsilon\to 0$, assume that $x$ is a Lebesgue point of $f$. Since $\rho$ is bounded, $$\begin{align*}\left|\int_{B(0,1)}\left(f(\epsilon z+x)-f(x)\right)(\rho(-z)-c)\,dz\right|&\leq C\int_{B(0,1)}|f(\epsilon z+x)-f(x)|\,dx\\
&=C\frac{1}{|B(x,\epsilon)|}\int_{B(x,\epsilon)}|f(y)-f(x)|\,dy,\end{align*}$$ and the last integral converges to $0$ from the definition of a Lebesgue point.
