Let $A$ be a compact subset of $\mathbb{R}$ of positive Lebesgue measure. Assume that almost everywhere the limit $f(x)=\lim_{\delta\to 0}\frac{\lambda(A\cap (x-\delta,x+\delta))}{2\delta}$ exists, where $\lambda$ is the Lebesgue measure.
How can I show that, for almost every $x\in A$, that $f(x)=1$?
You do not need $A$ to be compact, just positive measure will suffice. Neither you need the existence of a limit, it follows automatically from Lebesgue's theorem (see below).
Let $\mathbb{I}_A$ be the characteristic function of $A$. Then almost all (with respect to $\mu$) points of $\mathbb{R}$ are Lebesgue points of $\mathbb{I}_A$ in view of Lebesgue's differentiation theorem. Writing the condition of a point $x$ being a Lebesgue point for $\mathbb{I}_A$ means $$ \frac{1}{2\delta} \int\limits_{x - \delta}^{x + \delta} \mathbb{I}_A (y) dy \to \mathbb{I}_A(x) , \ \ \text{ as } \ \ \delta \to 0 , $$ which is precisely the limit in question.