Lebesgue point and integration Let $f$ be in $L^1_{\text{loc}}(\mathbb{R})$. We know that for almost every $t$
$$
\lim_{h\to 0} \frac{1}{h} \int_t^{t+h} |f(u)-f(t)|\text{d} u = 0.
$$
My question is : can we say that for almost every $s$
$$ 
\lim_{h\to 0} \int_0^s \frac{1}{h} \int_t^{t+h} |f(u)-f(t)|\text{d} u \text{d} t = 0 ?
$$
The quantity inside the integral (between $0$ and $s$) goes to $0$ but I can't find a bound independent of $h$ to apply dominated convergence.
 A: Let $h_{n}\downarrow0$ and define
$$
f_{n}:\left[0,s\right]\to\mathbb{R},t\mapsto\frac{1}{h_{n}}\int_{t}^{t+h_{n}}\left|f\left(u\right)-f\left(t\right)\right|\,{\rm d}u.
$$
By the Lebesgue differentiation theorem, $f_{n}\left(t\right)\xrightarrow[n\to\infty]{}0$
for almost all $t\in\left[0,s\right]$. Furthermore,
$$
0\leq f_{n}\left(t\right)\leq\frac{1}{h_{n}}\cdot\int_{t}^{t+h_{n}}\left|f\left(u\right)\right|\,{\rm d}u+\frac{1}{h_{n}}\int_{t}^{t+h_{n}}\left|f\left(t\right)\right|\,{\rm d}u=\frac{1}{h_{n}}\int_{t}^{t+h_{n}}\left|f\left(u\right)\right|\,{\rm d}u+\left|f\left(t\right)\right|=:g_{n}\left(t\right).
$$
Again, by the Lebesgue differentiation theorem, $g_{n}\left(t\right)\xrightarrow[n\to\infty]{}2\left|f\left(t\right)\right|$
for almost all $t\in\left[0,s\right]$.
Finally,
\begin{eqnarray*}
\int_{0}^{s}g_{n}\left(t\right)\,{\rm d}t & = & \int_{0}^{s}\left|f\left(t\right)\right|\,{\rm d}t+\frac{1}{h_{n}}\int_{0}^{s}\int_{t}^{t+h_{n}}\left|f\left(u\right)\right|\,{\rm d}u\,{\rm d}t\\
 & \overset{\text{Fubini}}{=} & \int_{0}^{s}\left|f\left(t\right)\right|\,{\rm d}t+\frac{1}{h_{n}}\int_{0}^{s+h_{n}}\int_{\max\{0, u-h_{n}\}}^{\min\{u,s\}}\left|f\left(u\right)\right|\,{\rm d}t\,{\rm d}u\\
 & = & \int_{0}^{s}\left|f\left(t\right)\right|\,{\rm d}t+\int_{0}^{s+h_{n}} \frac{\min\{u,s\} - \max\{0, u-h_{n}\}}{h_n} \cdot\left|f\left(u\right)\right|\,{\rm d}u\\
 & \xrightarrow[n\to\infty]{} & 2\int_{0}^{s}\left|f\left(t\right)\right|\,{\rm d}t=\int_{0}^{s}\lim_{n\to\infty}g_{n}\left(t\right)\,{\rm d}t.
\end{eqnarray*}
Hence, by a special version of the dominated convergence theorem (see
General Lebesgue Dominated Convergence Theorem),
we conclude
$$
\int_{0}^{s}f_{n}\left(t\right)\,{\rm d}t\xrightarrow[n\to\infty]{}0.
$$
EDIT: In the last step above, we used (for $n$ large enough):
\begin{eqnarray*}
 &  & \int_{0}^{s+h_{n}}\frac{\min\left\{ u,s\right\} -\max\left\{ 0,u-h_{n}\right\} }{h_{n}}\left|f\left(u\right)\right|\,{\rm d}u\\
 & = & \int_{h_{n}}^{s}\underbrace{\frac{u-\left(u-h_{n}\right)}{h_{n}}}_{=1}\left|f\left(u\right)\right|\,{\rm d}u+\int_{0}^{h_{n}}\underbrace{\frac{u-0}{h_{n}}}_{\in\left[0,1\right]}\left|f\left(u\right)\right|\,{\rm d}u+\int_{s}^{s+h_{n}}\underbrace{\frac{s-\left(u-h_{n}\right)}{h_{n}}}_{\in\left[0,1\right]}\cdot\left|f\left(u\right)\right|\,{\rm d}u\\
 & \xrightarrow[n\to\infty]{\text{dominated convergence (e.g.)}} & \int_{0}^{s}\left|f\left(u\right)\right|\,{\rm d}u
\end{eqnarray*}
