Suppose that $f:[a,b] \rightarrow \mathbb{R}$ is a right-continuous function, that is, $f(x+)=f(x)$ for all $x \in [a,b)$. Are these functions $L^1$?
I think that right-continuous functions would be (locally) bounded so that the answer would be affirmative, but I cannot prove this easily.
Maybe enumerating the rationals will allow me to construct an exotic counterexample? I'm not sure.
Best Answer
No, this is horribly false. For instance, on $[-1,0]$, you could have $f(x)=1/x$ (with $f(0)$ defined however you want). More generally, $f(x)$ could blow up as fast as you want as $x$ approaches some $c\in [a,b]$ from below, while being continuous above $c$.