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EDIT:

Perhaps a more reasonable question after thinking about the answer I got would have been.

Is there a set $N$ of measure $1-\varepsilon$ and a disjoint partition of that set $N$ with finitely many disjoint sets $I_i$ such that each of them contains an $x$ for which $$\left\lvert \left\lvert I_n \right\rvert^{-1} \int_{I_n} f(s) ds- f(x) \right\rvert \le \varepsilon?$$

END EDIT.

The differentiation theorem teaches us that for a $L^1_{loc}$ function

$$\frac{1}{B(x,R)}\int_{B(x,R)} f(s) ds \rightarrow f(x)$$

almost everywhere.

Now consider a Lebesgue integrable function on the interval $[0,1]$ and the partition $I_n^k=[k/n,(k+1)/n]$ for $k=0,..,n-1.$

Uniform rational case: I would like to ask whether for all $\varepsilon>0$ there exists a natural number $n$ such that for all $I_n^k$ there is an $x \in I_n^k$ such that

$$\left\lvert n\int_{I_n^k} f(s) ds- f(x) \right\rvert \le \varepsilon.$$

Uniform real case: Is the assumption of disjoint intervals $I_{\delta}^k$ of same size $\delta$ where $\delta>0$ is an arbitrary real number stronger?-Of course in this case, we do not always exactly cover $[0,1]$ but you may assume that the function actually lives on $[0,2]$ for convenience.

And even less restrictive:

Is there any finite disjoint partition of intervals $I_n$ of $[0,1]$ such that each of them contains an $x$ for which $$\left\lvert \left\lvert I_n \right\rvert^{-1} \int_{I_n} f(s) ds- f(x) \right\rvert \le \varepsilon?$$

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  • $\begingroup$ What are you asking $\exists\epsilon>0\forall n$ or $\forall n\exists\epsilon$? $\endgroup$ – Piotr Hajlasz Apr 22 '18 at 20:57
  • $\begingroup$ @PiotrHajlasz $\forall \varepsilon \exists n$. Sorry, I agree it was a written in an unclear way. $\endgroup$ – Sascha Apr 22 '18 at 20:59
  • $\begingroup$ Isn't the construct "disjoint sets such that each of them contains $x$" an oxymoron? What was really meant? $\endgroup$ – fedja Apr 23 '18 at 1:50
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The answer to all your questions is no.


Consider $f(x) = (-1)^n$ when $x \in (2^{-n-1}, 2^{-n}]$, $n = 0, 1, \ldots$ Suppose that $\delta \in (2^{-k-1}, 2^{-k}]$. The indefinite integral of $f$ is linear on $[2^{-k-1}, 2^{-k}]$ with values $(-2)^{-k-1}/3$ and $(-2)^{-k}/3$ on the endpoints. Therefore, $$\left|\int_0^\delta f(s) ds\right| \leqslant 2^{-k}/3 ,$$ and hence $$\left|\frac{1}{\delta} \int_0^\delta f(s) ds\right| \leqslant 2/3 .$$ On the other hand, $|f(x)| = 1$ in $[0, \delta]$. We conclude that $$\left|\frac{1}{\delta} \int_0^\delta f(s) ds - f(x)\right| \geqslant 1/3 $$ for every $x \in [0, \delta]$.

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    $\begingroup$ @Sascha: The function $f$ defined in the answer is a counter-example: $\epsilon^{-1} \int_0^\epsilon f(x) dx = (-1)^k/3$ for $\epsilon = 2^{-k}$. $\endgroup$ – Mateusz Kwaśnicki Apr 22 '18 at 21:27
  • $\begingroup$ @Sascha: I do not think I have much intuition here, sorry. Regarding the second part: indeed, you can easily cover $[0,1]$ up to a set of a given measure $\epsilon > 0$ with a greedy approach. $\endgroup$ – Mateusz Kwaśnicki Apr 22 '18 at 21:55

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