Measure Theory-Lebesgue Measurable

[WannaBe22]

Homework Statement

Let A \subseteq R be a Lebesgue-Measurable set. Prove that if the Lebesgue measure of A is less than infinity , then the function f(x) = \lambda(A \cap (-\infty,x)) is continous.

Homework Equations

The Attempt at a Solution

I'm really confused about the definition of \lambda (A) where \lambda is the Lebesgue-measure...I've tried taking an \epsilon >0 and choosing some \delta >0 for which if |x-x_0 | < \delta then |f(x)-f(x_0)| <\epsilon but I don't think this is the point...

I'll be delighted to get some guidance

Thanks !

 

[ystael]

Intuitively, f(x) is the measure of the portion of A "left of x". So if x < x', can you interpret f(x') - f(x) in terms of measures?

 

[WannaBe22]

Intuitively, f(x')- f(x) is measure of the portion of A between x and x' ... Intuitively , this whole thing seems quite trivial...But when I try to get to the formal aspect of the soloution (as seen in "The attempt at a solution" part) , everything messes out... How can I make the intuition more formal ?
I really hope you'll be able to help me

Thanks !

 

[ystael]

Here are two hints:

1. Given that intuitive description of f(x') - f(x), try to come up with an upper bound for f(x') - f(x) in terms of x' - x. This is what you need to prove continuity. (What property of A would give the largest possible value for f(x') - f(x)?)

2. The fact that \lambda(A) < \infty is irrelevant to the continuity of f; you need it merely to define f.

 

[micromass]

What is f(x)-f(x₀)?

Hint: you must use that if A\subseteq B and if those two sets have finite measure, then \lambda(B\setminus A)=\lambda(B)-\lambda(A).

 

[WannaBe22]

Thanks a lot! your guidance was very helpful!