Fact in proof of Lebesgue's Differentiation Theorem I'm reading a proof of Lebesgue's Differentiation Theorem, where there is a fact that is not further specified. Let $f \in L^1(\mathbb R^n)$. For $r > 0$ we set 
$$f_r(x) := \frac{1}{\lambda^n(\mathbb B(x, r))} \int_{\mathbb B(x, r)} f(y) \, \mathrm dy \; .$$
Now the author states, that 
$$ \Vert f_r - f \Vert_{L^1(\mathbb R^n)} \to 0 \quad \text{for } r \to 0 \; .$$
Why does this hold? Is it obvious?
 A: Yes, this is true. One way to see this is to note that
$$
f_r = g_r \ast f,
$$
with $g_r = \frac{1}{\lambda^n (B_r (x))} \cdot \chi_{B_r (x)}$, where $\ast$ denotes the convolution.
Since the family $(g_r)_r$ is an approximation of unity, this implies $f_r \to f$ in $L^p$ for every $1\leq p<\infty$ for which $f \in L^p$ is true.
If you do not know that much about convolutions, I can elaborate on this answer.
Ok, here come the details: The following requires you to know that $\Vert L_x f - f\Vert_1 \to 0$ as $x \to 0$. This is for example proved here (proof that translation of a function converges to function in $L^1$), but the proof requires knowing that $C_c$ functions are dense in $L^1$.
We let $(g_\epsilon)_\epsilon$ be any family of $L^1$ functions with $\int |g_\epsilon| \, dx \leq C$ and $\int g_\epsilon \, dx = 1$ for all $\epsilon$. Finally, we assume ${\rm supp} g_\epsilon \subset \overline{B_\epsilon (0)}$. It is easy to see that the family $(g_r)_r$ defined above satisfies this.
Now,
\begin{eqnarray*}
\left\Vert \left(g_{\epsilon}\ast f\right)-f\right\Vert _{L^{1}} & = & \int\left|f\left(x\right)-\int g_{\varepsilon}\left(y\right)\cdot f\left(y-x\right)\, dy\right|\, dx\\
 & \overset{\int g_{\epsilon}\left(y\right)\, dy=1}{=} & \int\left|\int g_{\epsilon}\left(y\right)\cdot\left[f\left(x-y\right)-f\left(x\right)\right]\, dy\right|\, dx\\
 & \leq & \int\int\left|g_{\epsilon}\left(y\right)\right|\cdot\left|f\left(x-y\right)-f\left(x\right)\right|\, dy\, dx\\
 & \overset{\text{Fubini}}{=} & \int\int\left|g_{\epsilon}\left(y\right)\right|\cdot\left|f\left(x-y\right)-f\left(x\right)\right|\, dx\, dy\\
 & = & \int\left|g_{\epsilon}\left(y\right)\right|\cdot\left\Vert f-L_{y}f\right\Vert _{L^{1}}\, dy\\
 & \overset{{\rm supp}\, g_{\epsilon}\subset\overline{B_{\epsilon}\left(0\right)}}{\leq} & \int\left|g_{\epsilon}\left(y\right)\right|\, dy\cdot\sup_{y\in\overline{B_{\epsilon}\left(0\right)}}\left\Vert f-L_{y}f\right\Vert _{L^{1}}\\
 & \leq & C\cdot\sup_{y\in\overline{B_{\epsilon}\left(0\right)}}\left\Vert f-L_{y}f\right\Vert _{L^{1}}\xrightarrow[\varepsilon\downarrow0]{}0.
\end{eqnarray*}
One way to see that $C_c$ is dense in $L^1$ is to use regularity of the Lebesgue measure: Since the simple functions are dense, it suffices to find for each measurable set $M \subset \Bbb{R}$ with finite measure some $g \in C_c$ with $\Vert \chi_M - g\Vert_1 < \epsilon$. But be regularity, there are $K \subset M \subset U$ with $K$ compact, $U$ open and $\lambda(U \setminus K) < \epsilon$.
Using (e.g.) Urysohns Lemma, we find a function $g \in C_c$ with ${\rm supp}g \subset U$ and $g \equiv 1$ on $K$. It is then easy to see $\Vert \chi_M - g\Vert_1 < \epsilon$.
This proof is suitable here, because the other standard method of showing that $C_c$ is dense already uses the above claim that $g_r \ast f \to f$ for a suitable family $(g_r)_r$.
