Following wikipedia, we know that the Lebesgue differentiation theorem holds for quite general measures on separable metric spaces as long as they are finite dimensional, more explicitly if the measure is doubling, or the metric space is a Riemannian manifold or if the metric space is locally compact and the metric is an ultrametric.
What I gather is that there can exist a Banach $X$, a positive measure $\mu$ and a measurable function $f:X\to \mathbb R$ such that $$ \mu\left\{x\in X \,\middle|\,\frac{1}{\mu(B(x,\varepsilon))}\int_{B(x,\varepsilon)}f\mu\not\to f(x)\right\}>0. $$
I understand that one standard proof of this theorem relies on the density of continuous functions inside the integrable ones and I read somewhere that there can be continuous functions which are not locally integrable. Sadly I managed only to find an example of a continuous curve inside a Topological Vector Space which isn't integrable, while what I expect here is a real valued continuous function which isn't integrable.
Question 1) What fails in the proof of the Lebesgue differentiation theorem for a general measure in a separable metric space?
Question 2) do you know of a space where the continuous functions with compact support are not dense in the integrable functions for a given measure?