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I would be very thankful if someone could answer this question. I am new here so I copied a similar question and I asked my doubt.

$$\lambda(A) = \lambda (A \cap E) + \lambda (A \cap E^c)$$

where $\lambda$ is an outer measure, $A \subset R$, $E \subset R$, and $E$ is an elementary set. That is, $E$ is a union of finitely many bounded intervals.

By countable sub-additivity, I know

$$ \lambda(A) \leq \lambda(A \cap E) + \lambda(A \backslash E) $$

since $A = (A \cap E)\cup (A \backslash E)$.

I need help understanding the opposite inequality.

By the definition of the Lebesgue outer measure, for each $\epsilon>0$, there an exists open set $O$, with $O \supset A$ such that $\lambda(O) \leq \lambda(A) + \epsilon$. Now observe that

$$ \lambda(A) + \epsilon \geq \lambda(O) = \lambda(O \cap E) + \lambda (O \backslash E) \geq \lambda(A \cap E) + \lambda (A \backslash E) $$

The equality above comes from the fact that the Lebesgue outer measure is countably (and finitely) additive for Lebesgue measurable sets.

Since $\epsilon$ is arbitrary, we have $\lambda (A) \geq \lambda(A \cap E) + \lambda(A \backslash E)$.

I do not understand the role of $\epsilon$. How can $\lambda(A) \geq \lambda(A \cap E) + \lambda(A \backslash E)$ when $\epsilon$ is zero?

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This is a basic argument from real analysis/advanced calculus. The chain of inequalities you derived gives you that, for all $\epsilon > 0$,

$$ \lambda(A) + \epsilon \geq \lambda(A \cap E) + \lambda(A \cap E^c) $$

Since this is true for all $\epsilon > 0$, it must be true for $\epsilon = 0$ as well.

Suppose not. Then:

$$ \lambda(A) < \lambda(A \cap E) + \lambda(A \cap E^c) $$

which means we can find $\epsilon$ small enough so that

$$ \lambda(A) + \epsilon < \lambda(A \cap E) + \lambda(A \cap E^c) $$

which is a contradiction.

Alternatively, recall that, in $\mathbb R$, limits preserve inequalities. (That being said, I suspect you'll need to employ an argument by contradiction similar to the one I just gave to prove this fact if you don't already know it.)

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  • $\begingroup$ Thanks for all the quick answers, and sorry for bothering. The thing is I do not know if I have already understood the concept of outer measure. I would like to ask if we have a A whose outer measure is higher than B. Does it mean B is going to use for its outer measure less elements of the covering of A. If both are contained in the union of the same elements? $\endgroup$ – Pedro Gomes Dec 27 '16 at 19:06
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The choice of open set $O$ depends on your $\epsilon$. You initially use $O$ to deduce some inequalities but then you get an inequality involving $\epsilon$ but no $O$. So the $\epsilon$ dependence is now missing, as a result the inequality is true for all $\epsilon>0$. Take $\epsilon\to0$ to get the answer finally.

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  • $\begingroup$ I truly appreciate your answer. However I may not be understanding this fully. Once epsilon goes to 0, lambda of A gets lower than lambda O. Or does Lambda A is equal to Lambda O? $\endgroup$ – Pedro Gomes Dec 20 '16 at 13:18

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