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Let $F:[0,1] \to \mathbb{R}$ be a function such that $F'(x)$ exists almost everywhere in $[0,1]$, $F'$ is Lebesgue integrable in $[0,1]$, $F$ is continuous at $x = 0$ and $F$ is absolutely continuous on $[\epsilon,1]$ for every $\epsilon > 0$.

Prove $F$ is absolutely continous on $[0,1]$.

I found some posts here assuming $F$ is continous, but in this case I have only continuity at $x =0$, so I don't quite know what to do.

I tried breaking $[0,1]$ into $[0,\epsilon) \cup [\epsilon,1]$, so that it is enough proving ac on $[0,\epsilon)$. Then I tried using the Lebesgue differentiation Theorem for $F'$, since it is $L^1([0,1])$, but couldn't properly write something that helped with absolutely continuity...

Arctic Char
  • 16,007

1 Answers1

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We use the fact that $f: [0,1] \to \mathbb R$ is absolute continuous $\Leftrightarrow$ $f$ is differentiable almost everywhere, $f'$ is Lebesgue integrable and for all $0\le x<y\le 1$,

$$\tag{1} f(y) - f(x) = \int_x^y f'(s) ds.$$

Since $f$ is absolutely continuous in $[\epsilon, 1]$ for all $\epsilon >0$, (1) holds when $x>0$. The case $x=0$ follows by taking $x\to 0$ in (1): Since $f$ is continuous at $x=0$, $$\lim_{x\to 0^+} (f(y) - f(x)) = f(y)-f(0).$$

Since $f'$ is integrable, write $\int_x^y f'(s)ds = \int_0^y \chi_{[x, y]} f'(s) ds$, then

$$\lim_{x\to 0^+} \int_x^y f'(s)ds = \int_0^y f'(s) ds$$ follows by Lebesgue dominated convergence theorem. Thus $(1)$ holds also for $x=0$ and thus $f$ is absolutely continuous in $[0,1]$.

Arctic Char
  • 16,007