Continuity of Integrals in L^1 Spaces

[Oxymoron]

Question:

Prove that if $f \in L^1(\mathbb{R},\mathcal{B},m)$ and $a \in \mathbb{R}$ is fixed, then $F(x):=\int_{[a,x]}f\mbox{d}m$ is continuous. Where $\mathcal{B}$ is the Borel $\sigma$-algebra, and $m$ is a measure.

 

[Oxymoron]

I was hoping to use the following definition:

A function $f$ is continuous if for any sequence $x_n$ such that

$$x_n \rightarrow x$$

then

$$F(x_n) \rightarrow F(x)$$

Does this sound like the right approach?

 

[StatusX]

Maybe. Can you show that $\int_a^b+\int_b^c=\int_a^c$, and then that as x_n->x, $\int_{x_n}^x f dm$->0?

 

[Oxymoron]

Im not sure that would help. But what do I know!?

I was thinking that to show that F(x) was continuous I would do something like this:

1) Fix $a \in \mathbb{R}$ and let $x_n$ be a sequence that converges to x as n approaches infinity. You know, all the regular proof setting up stuff.

2) Use the Dominated Convergence Theorem to show that the sequence $F(x_n)$ converges, and finally get something like

3) $$\int_{[a,x]}f_n\mbox{d}m = \lim_{n\rightarrow\infty}\int_{[a,x]}f_n\mbox{d}m$$

Hence showing that

$$F(x_n) \rightarrow_{n\rightarrow\infty} F(x)$$

So basically I think using the D.C.T. is essential here. What does anyone think of this method?

 

[StatusX]

I'm guessing the f_n are functions that are equal to f everywhere except in (x_n,x), where they are 0. So you need to show that the integral from [a,x_n] of f is equal to the integral over [a,x] of f_n, and that the integrals of the f_n converges to the integral of their limit, f, which you can do using the dominated convergence theorem, bounding the |f_n| above by |f|. Sounds good. My suggestion was just to show that the error (the integral over [x_n,x] of f) goes to zero as x_n goes to x.