I'm studying the Lebesgue version of the Fundamental Theorem of Calculus, and I came across this theorem, which embodies the counterpart of TFC's first part
Given $f\in L^1(\mathbb{R})$, we have that $$\frac{d}{d x} \int_{x_0}^x f(t) dt = f(x) \quad \text{a.e.}$$
Now I'm asking if this theorem can be extended in the following way:
Given $f\in L_{loc}^1(\mathbb{R})$, we have that $$\frac{d}{d x} \int_{x_0}^x f(t) dt = f(x) \quad \text{a.e.}$$
The proof I've seen of the first theorem uses in a essential way the integrability of the function (over $\mathbb{R}$ ).
The indefinite Lebesgue integral of the locally integrable function is differentiable a.e. by the Lebesgue theorem on monotone functions. The question therefore is whether the equality holds (a.e.). So:
- Can the theorem be proved for $L^1_{loc}(\mathbb{R})$ functions?
If not:
What can say about that derivative (which exists a.e.)?
Is there a counterexample?
Thanks
Yes, the Lebesgue differentiation theorem holds for any locally integrable function. Just note that if $f\in L^1_{\text{ loc }}(\Bbb R )$ then the compactness of $[-n,n]$ show us that $f|_{[-n,n]}\in L^1([-n,n])$ for every $n \in \Bbb N $. Then $$ \frac{\mathrm{d}}{\mathrm{d}y}\int_x^y f(t)\,\mathrm d t=f(y)\tag1 $$ for almost every $y\in (-n ,n )$ and any chosen $x\in(-n,n)$. Let $N_n$ be the null set where the identity of $\rm(1)$ doesn't holds for $n\in \Bbb N $, then the sequence $(N_n)$ is increasing and so $$ \lambda \left(\bigcup_{n\in \Bbb N }N_n\right)=\lim_{n\to \infty }\lambda (N_n)=0\tag2 $$ where $\lambda $ is the Lebesgue measure. Then $\rm(1)$ holds for any chosen $x\in \Bbb R $ and almost every $y\in \Bbb R $.