Lebesgue measurable characterisation proof

Asked Dec 18 '13 at 17:52

Active Oct 29 '21 at 23:42

Viewed 333 times

Let $I$ be bounded subinterval of $\mathbb{R}$. Show that a $B\subseteq I$ is Lebesgue measurable if and only if it satisfies $$\lambda^*(I)=\lambda^*(B)+\lambda^*(I\cap B^c)$$

First implications goes from definition of Lebesgue measurability where we just take $I$ as test set.

But I'm stuck with second implication. I tried to prove it by definition:

Let $A\subseteq\mathbb{R}$ be arbitrary set. We need to prove $\lambda^*(A) \ge \lambda^*(A \cap B) +\lambda^*(A \cap B^c) $ (other inequality goes from sigma aditivity of outer measure). And I can't even see how to use statement from problem to prove this. Any help?

edited Oct 29 '21 at 23:42

KCd

46,062

asked Dec 18 '13 at 17:52

Meow

1,910

Use the seqrch over the site. Defenitely this was answered earlier – Norbert Dec 18 '13 at 18:32
Couldn't find it. I know proof if statement is true for every interval $I$ but not for this one... – Meow Dec 19 '13 at 08:51
1

What about this? – Norbert Dec 19 '13 at 10:22
Hmm... but I am given first interval $I$ and then any subset. What if $\lambda^*(I-B)\neq 0$. How can I use that characterization? – Meow Dec 19 '13 at 11:00

Lebesgue measurable characterisation proof

0 Answers0