Show Continuity of F(x) in Lebesgue Integrable f [a,b]

[math8]

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] )

 

[quasar987]

Write

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

for h>0, and

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

for h>0, and use Lebesgue Dominated Convergence theorem to show that F(x+h)-F(x)-->0.

 

[Hurkyl]

That's an interesting thought. I would have considered

F(x+h) - F(x) = \int_x^{x+h} f(t) \, dt

 

[math8]

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?

 

[gsenel]

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]

Well, make it an exercice to show that for any x in [a,b],

\chi_{(x,x+h)}\rightarrow\chi_{\emptyset}

(\chi_{\emptyset} 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 1\leq r\leq s\leq +\infty, L^s(X)\subset L^r(X)[/tex]. Because then you can just write<br /> <br /> |F(x+h)-F(x)|\leq \int_x^{x+h}|f(t)|dt\leq||f||_{\infty}(x+h-x)\rightarrow 0

 

[Hurkyl]

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]

Scratch what I said about Hurkyl's idea.