Hi in the proof of f integrable, $g(x)= \int_{[0,x]} f dt$ is uniformly continuous, i saw a proof that uses the following argument. Let $\epsilon > 0$ be given, take $δ := \epsilon^-1 · (\int |f|dt) >0$ since f is integrable. Let $x<y $such that $|x − y| < δ$, Then,$|g(x) − g(y)| ≤ \int_{[x,y]} |f(t)|dt ≤ |x − y|\int_{R} |f(t)|dt < \epsilon. $
My question is, how do we get that $\int_{[x,y]} |f(t)|dt ≤ |x − y|\int_{R} |f(t)|dt$? I have tried to bound it and only able to conclude that $\int_{[x,y]} |f(t)|dt\leq |x − y| sup(|f|)$ ?
Thanks in advance
You can't, the proposed proof is wrong. A simple counterexample is $f(x) = |x|^{-1/2}$ for $-1 \le x \le 1$, $f(x)=0$ otherwise. Note $g(x) = 2 \sqrt{|x|}$ for $-1 \le x \le 1$ which is not Lipschitz.