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Is it true that if $A \subseteq \mathbb{R}$ is Lebesgue measurable set, then

i) the set $A/2$ is Lebesgue measurable,

ii) the set $-A$ is Lebesgue measurable.

The definition I have for Lebesgue measurability of a set $A$ is: Let $\lambda^*()$ be the Lebesgue outer measure. A set $A$ is called Lebesgue measurable if $\lambda^*(X) = \lambda^*(X \cap A) + \lambda^*(X \cap A^c)$, for any arbitrary set $X$.

I think both are true but I can't show them. A proof sketch or a counter example would be very helpful. Thanks!

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  • $\begingroup$ What definition do you have for Lebesgue measurability of a set? I would like you to state it, so that there is no confusion if somebody attempts an answer and your definitions clash or don't obviously agree. $\endgroup$ – Teresa Lisbon Mar 29 '18 at 23:50
  • $\begingroup$ If $\lambda^*()$ is the Lebesgue outer measure. A set A is measurable if $\lambda^*(X) = \lambda^*(X \cap A) + \lambda^*(X \cap A^c)$, for any arbitrary set $X$. $\endgroup$ – Yahia Shabara Mar 29 '18 at 23:54
  • $\begingroup$ Suppose you are working with $-A$. Fix any subset $X$. Can you write $X \cap -A$ as something intersected with $A$, maybe translated, etc.? You can : note that $$X \cap -A = \{x : x \in X, -x \in A\} = \{-y : -y \in Y, y \in A\} = -(-X \cap A)$$. Now, do something similar with $A^c$, and try using the fact that $A$ is Lebesgue measurable. $\endgroup$ – Teresa Lisbon Mar 30 '18 at 0:01
  • $\begingroup$ Thanks a lot for your help!.. There is still one step missing for me which (I guess) is showing that $\lambda(X \cap A) = \lambda(- (-X \cap A)) $. $\endgroup$ – Yahia Shabara Mar 30 '18 at 0:25
  • $\begingroup$ If you have a set of rectangles(or boxes, or whatever you have in the definition of outer measure) covering one of those, you have a similar set of boxes covering the other one, which has the same area. Prove this (easy) and then the above statement is obvious. $\endgroup$ – Teresa Lisbon Mar 30 '18 at 0:27
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(Literally copy pasted)

  • Let us say you are working with $-A$. Fix any $X$, and note that $X \cap -A$ = $-(-X \cap A)$. You can do something similar with $A^c$.

  • To use the fact that $A$ is Lebesgue measurable, one correlates covers by rectangles of $-X \cap A$ with covers of $X \cap -A$. This is because they are the same set up to translation.

Now, for $\frac A2$, maybe you will find some other relation like $X \cap \frac{A}{2} = ? \cap A$, and something similar for $A^c$, and the same sort of tricks will apply.

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