In the question, $\lambda_{n} (P)$ is Lebesgue measure on Lebesgue $\sigma$-algebra on $\mathbb{R^n}$. This is probably very easy, but i'm stuck. I know that $\lambda_{n} (A) + \lambda_{n} (P \setminus A) = \lambda_{n} (P)$ and it follows that $P \setminus A$ is a null set. How do I prove that it is empty?
Prove that if $P \subset \mathbb{R^n}$ is a closed interval, $A \subseteq P$ is a closed set and $\lambda_{n} (P) = \lambda_{n} (A)$ then $P=A$.
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measure-theory
lebesgue-measure
2 Answers
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Let $Q$ denote the interior of $P$. Note that $\lambda_n(P \setminus Q) = 0$.
You have $$\lambda_n(A) = \lambda_n(P) = \lambda_n(Q) = \lambda_n(Q \cap A) + \lambda_n(Q \setminus A)$$ and $$\lambda_n(Q \cap A) = \lambda_n(P \cap A) = \lambda_n(A)$$ from which you get $\lambda_n(Q \setminus A) = 0$. Since $Q \setminus A$ is open, it follows that $Q \setminus A = \emptyset$. Thus $A \cap Q = Q$ so that $$P = \overline Q = \overline{A \cap Q} \subset \overline A = A \subset P.$$ Thus $A=P$.
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I will try to show that $A$ is dense in $P$ with the induced topology. Since $A$ is closed in $P$, $P=A$.
Let $B\cap P$, $B$ open in $\mathbb {R}^n$, be an open set in the induced topology.
If $B\cap P\neq \emptyset$ I notice that $B \cap P$ has at least an interior point (hint, if the intersection has only border points of P, $B$ is not open, has some non interior points, so it has to be $B \cap int(P) \neq \emptyset$, which is an open set. Here I'm using that $P$ is an interval)
$\lambda(B\cap P) \geq \lambda(B\cap int(P))> 0 $ since this set is open in $\mathbb{R}^n$ and has to have inside an open interval of the basis.
This means that $B \cap P \not\subset P\setminus A$ because of monotonicity of measure, so every open in $P$ has to intersecate with $A$. So $A$ is dense in $P$.