Continuity of F(x)

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Let F(x) = Integral from a to x of f dt (a belongs in [a,b])
How do we show that F(x) is continuous? (f is Lebesgue integrable on [a,b] )
 

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quasar987
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Write

[tex]F(x+h)-F(x)=\int_a^b\chi_{(x,x+h)}f(t)dt[/tex]

for h>0, and

[tex]F(x+h)-F(x)=-\int_a^b\chi_{(x+h,x)}f(t)dt[/tex]

for h>0, and use Lebesgue Dominated Convergence theorem to show that F(x+h)-F(x)-->0.
 
Hurkyl
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That's an interesting thought. I would have considered

[tex]F(x+h) - F(x) = \int_x^{x+h} f(t) \, dt[/tex]
 
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quasar987, thanks but I don't see how the DCT shows that F(x+h)-F(x)-->0. I know that
lim integral =integal lim
but does lim K(x+h,x)*f(t) tends to 0? If yes, why?
 
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quasar987, thanks but I don't see how the DCT shows that F(x+h)-F(x)-->0. I know that
lim integral =integal lim
but does lim K(x+h,x)*f(t) tends to 0? If yes, why?
As far as I understand the limit of the indicator function will tend to 0 which in return will make the integral 0.
 
quasar987
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Well, make it an exercice to show that for any x in [a,b],

[tex]\chi_{(x,x+h)}\rightarrow\chi_{\emptyset}[/tex]

([itex]\chi_{\emptyset}[/itex] is just the function that is identically 0).

In a way, Hurkyl's way is swifter if you know that in a finite measure space X (such as [a,b] with the Lebesgue measure), for any [itex]1\leq r\leq s\leq +\infty[/itex], [itex]L^s(X)\subset L^r(X)[/tex]. Because then you can just write

[tex]|F(x+h)-F(x)|\leq \int_x^{x+h}|f(t)|dt\leq||f||_{\infty}(x+h-x)\rightarrow 0[/tex]
 
Hurkyl
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Well, I was just thinking about the mean value theorem for integrals. But either way, things become a little trickier when f isn't bounded near x....
 
quasar987
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Scratch what I said about Hurkyl's idea.
 

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