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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 http://math.stackexchange.com/questions/40384/showing-uniform-continuity?lq=1, http://math.stackexchange.com/questions/145222/f-is-integrable-prove-fx-int-inftyx-ft-dt-is-uniformly-continuo?lq=1, http://math.stackexchange.com/q/82862/. $F$ is Lipschitz if and only if $f$ is essentially bounded. It is not typically true that $|f(x)|\leq \|f\|_1$ a.e.2012-10-24
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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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    How do you justify the inequality?2012-10-24
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    The last inequality is false for general $L_1$ functions.2012-10-24
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    $F$ may not be Lipschitz, but still, is it continuous?2012-10-24
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    @LukasGeyer why? The domain is bounded, i just used Holder.2012-10-24
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    Holder would give $\int |f 1_{[x,y]}| \leq \|f\|_1 \|1_{[x,y]}\|_\infty$ not the bound you have...2012-10-24
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    Exactly, Hölder's inequality (actually, the easy limit case with conjugate exponents $1$ and $\infty$) would give you $\le \|f\|_\infty |x-y|$. This shows that you get Lipschitz continuity if $f \in L^\infty$. To see that you don't get it in $L_1$, just use $f(x) = x^{-1/2}$ on $[0,1]$, then $F(x) = 2\sqrt{x}$ which is only $1/2$-Hölder at $0$.2012-10-24
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    @LukasGeyer is it right now?2012-10-24
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    Is it right now @copper.hat?2012-10-24
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    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.2012-10-24
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    @copper.hat , interesting. I like your answer.2012-10-24
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    Yeah, looks good now.2012-10-24