If $f \in L^1([a, 1])$ for all $a \in (0, 1)$, is it true that $f \in L^1((0, 1])$?

Question

I'm learning about measure theory and need help with the following question:

True or Fasle (justify): If $f \in L^1([a, 1])$ for all $a \in (0, 1)$, then $f \in L^1((0, 1])$.

While it is very tempting to just substitute $a=0$ and conclude that the statement is true this is certainly the wrong way to go since $0 \notin (0, 1)$.

This problem got me thinking of the function $f(x) = {1 \over x}$. The latter function satisfies the conditions of the problem; the integral converges on any closed interval of the form $[a, 1]$ that does not contain $0$. So I have found an example that satisfies the given statement. This is not enough to conclude that the statement is true though (and I could not find any counterexample to disprove it). How should I proceed here?

Answer 1

Since $f(x) = \dfrac 1x$ does not belong to $L^1((0,1])$, your example shows that the statement is false.