Let A &#x2282;<!-- ⊂ --> [ a , b ) </mrow> for &#x2212;

Dayami Rose

Dayami Rose

Answered question

2022-06-24

Let A [ a , b ) for < a < b < be Lebesgue measurable. Show, that F ( x ) = μ ( [ a , x ) A ) for a x b is continous on [ a , b ]
I honestly don't know where to start here. My idea would be to somehow argue that because μ is continuous, then F must be continuous too. But this seems too easy.
Another idea I'd have is to do the following: Because we know, that A [ a , b ) , this means, that
[ a , x ) A = [ a , y ]
for some a y x. This means, that we can instead of F ( x ) use
G ( y ) = μ ( [ a , y ] )
G 1 ( μ ( [ a , y ] ) ) = [ a , x ]
So a closed image gives you a closed pre-image, which is one of the conditions for G to be continous. And if G is continuous, then F is, too.
I kind of know that my attempt is not good, because I assume that we should do it in a completely different way.

Answer & Explanation

Lilliana Burton

Lilliana Burton

Beginner2022-06-25Added 19 answers

The intersection is not necessarily an interval. You can use continuity of measure though, take h 0, then for x ( a , b )
| μ ( A [ a , x ) ) μ ( A [ a , x + h ) ) | = μ ( A [ x , x + h ) ) μ ( [ x , x + h ) μ ( { x } ) = 0.
and similarly for h < 0.
Continuity at the endpoints follows similarly. I leave those details to you.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?