Help with continuity of integral function

Question

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...

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. — Jonas Meyer, Oct 24 '12 at 15:27

Tomás · Accepted Answer · 2012-10-24T17:21:28.313

2

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

edited Oct 24 '12 at 17:21

answered Oct 24 '12 at 15:50

Tomás

22,559

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
Holder would give $\int |f 1_{[x,y]}| \leq |f|1 |1{[x,y]}|_\infty$ not the bound you have... – copper.hat Oct 24 '12 at 16:20
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 $0$. – Lukas Geyer Oct 24 '12 at 16:25
@LukasGeyer is it right now? – Tomás Oct 24 '12 at 16:55
Is it right now @copper.hat? – Tomás Oct 24 '12 at 17:57
1

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
@copper.hat , interesting. I like your answer. – Tomás Oct 24 '12 at 18:21
Yeah, looks good now. – Lukas Geyer Oct 24 '12 at 18:42

Help with continuity of integral function

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