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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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    I think the answer is right now @Tanya2012-10-24

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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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    Yeah, looks good now.2012-10-24