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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.

user37238
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  • Your first equation is true for every $t$, isn't it? (And "almost every $s$" makes no sense for your second equation, because it's about intervals $(0,s)$.) – TonyK Nov 24 '14 at 13:01
  • @TonyK The first equation is true only for the Lebesgue points of $f$, so no the first equation isn't true for every $t$. For the second equation, $s$ belongs to $\mathbb{R}$, so my "almost every" $s$ means that for almost every $s$ with respect to the Lebesgue measure on $\mathbb{R}$. – user37238 Nov 24 '14 at 14:10

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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*}

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PhoemueX
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  • I think that the change of variable, change the integral between $0$ and $s$ and the integral between $t$ and $t+h_n$ into an integral between $h_n$ to $s+h_n$ and an integral between $u-h_n$ and $u$, right? Otherwise, it is perfect. Thanks. – user37238 Nov 24 '14 at 14:12
  • I am not sure that this is correct. For example in the first line, if $t = 0$, then $u$ ranges from $[t, t+h_n] = [0,h_n]$ and if $t = s$, then $u$ ranges from $[t,t+h_n] = [s,s+h_n]$. So all in all, $u$ can range from $0$ to $s+h_n$. What you do here is (e.g.) to reformulate the integral boundaries as characteristic functions, i.e. $\int_{u-h_n}^u ... , dt= \int \chi_{[u-h_n, u]}(t) , dt$. Then use Fubini and compute the new boundaries from the characteristic functions. – PhoemueX Nov 24 '14 at 14:16
  • Your domain ${ 0\le u \le s+h_n, u-h_n\le t \le u }$ contains the point $(0,-h_n)$ that doesn't belong to ${ 0 \le t \le s, t \le u \le t+h_n }$, right? – user37238 Nov 24 '14 at 14:27
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    Okay, I adapted my post, I hope it is correct now. The last step (where the convergence is claimed) now requires a bit more thought, but is not too hard to see. Thanks for the correction. – PhoemueX Nov 24 '14 at 14:53