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Let $f\in L^1([0,1],\lambda)$ I'd like to show that $F(x)=\int_{[0,x]}|f|\, d\lambda$ is continuous.

I'm thinking of showing it is Lipschitz, but I can't really find any upper bound for $f$. Or maybe I can say something like $|f(x)|\leq \|f\|_1$ almost everywhere?

Any help is welcome...

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See…,…, $F$ is Lipschitz if and only if $f$ is essentially bounded. It is not typically true that $|f(x)|\leq \|f\|_1$ a.e. – Jonas Meyer Oct 24 '12 at 15:27
I think the answer is right now @Tanya – Tomás Oct 24 '12 at 17:57
up vote 2 down vote accepted

Let me try again. Suppose $x_n\rightarrow x$. Let $\Phi_{Y}$ be the characteristic function on $Y$. Note $$|f(y)|\Phi_{[0,x_n]}\rightarrow |f(y)|\Phi_{[0,x]},\ a.e.\ y\in[0,1]$$ and $$|f(y)|\Phi_{[0,x_n]}\leq|f(y)|,\ y\in[0,1]$$

Hence, by Lebesgue theorem we have $$\int_0^1|f(y)|\Phi_{[0,x_n]}\rightarrow \int_0^1|f(y)|\Phi_{[0,x]}$$

or equivalently $$\int_0^{x_n}|f(y)|\rightarrow \int_0^x|f(y)|$$

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How do you justify the inequality? – Tanya Oct 24 '12 at 15:58
The last inequality is false for general $L_1$ functions. – Lukas Geyer Oct 24 '12 at 16:05
$F$ may not be Lipschitz, but still, is it continuous? – Tanya Oct 24 '12 at 16:08
@LukasGeyer why? The domain is bounded, i just used Holder. – Tomás Oct 24 '12 at 16:10
Looks good. Another way would be to notice that the measure $\phi(A) = \int_A |f| d \lambda$ is absolutely continuous. Then if $\lambda A$ is small enough, $\phi A$ will be small. – copper.hat Oct 24 '12 at 18:14

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