**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 $$\left\lvert \left\lvert I_n \right\rvert^{-1} \int_{I_n} f(s) ds- f(x) \right\rvert \le \varepsilon?$$