Precise statement of the Lebesgue differentiation theorem

Question

Let

• $\lambda$ denote the Lebesgue measure on $\mathbb R$
• $T>0$ and $I:=(0,T)$
• $E$ be a $\mathbb R$-Banach space
• $f\in L^1(I,E)$

The most cited version of the Lebesgue differentiation theorem yields that $$\tilde f(t)=\lim_{h\to0}\frac1h\int_t^{t+h}f(s)\:{\rm d}s\;\;\;\text{for }\lambda\text{-almost all }t\in I\tag1$$ for any representative $\tilde f$ of $f$.

Now, I'm curious whether

1. the limit on the right-hand side exists for all $t\in I$. In that case, the statement could be rephrased in the following way: $f$ has a representative $\tilde f$ with $$\tilde f(t)=\lim_{h\to0}\frac1h\int_t^{t+h}f(s)\:{\rm d}s\;\;\;\text{for all }t\in I\;.\tag2$$
2. the statement can even been shown, if $f$ is only in $L_{\text{loc}}^1(I,E)$.

Answer 1

1. No, the limit may fail to exist. For one example, $f(x)=1/\sqrt{|x|}$ has infinite limit of averages as $x\to 0$. If infinite limits are acceptable, there are still counterexamples: let $f(x)=\sum_{n=0}^\infty \chi_{I_k}$ where $I_k = [2^{-2k}, 2^{1-2k}]$. This is a function that alternates between $0$ and $1$. Its average on the interval $[0, 2^{-2k}]$ is $$2^{2k}\sum_{j={k+1}}^\infty 2^{-2j}=2^{2k}\frac{2^{-2(k+1)}}{1-1/4} = \frac{1}{3}$$ while the average on $[0, 2^{1-2k}]$ is $$2^{2k-1}\sum_{j={k}}^\infty 2^{-2j}=2^{2k-1}\frac{2^{-2k}}{1-1/4} = \frac{2}{3}$$
2. Whatever is true for $L^1$ functions is true for $L^1_{loc}$ functions, because the statement "property holds a.e." is local in nature. If every point has a neighborhood where the property holds, then it holds everywhere (as long as we are on a second countable space, such as an interval, where a countable union of neighborhoods is enough).