How Can You Approximate a Borel Set with an Open Set in Measure Theory?

[jeterfan]

Homework Statement

We're given the measure space $$(X,A,μ)$$ with $$X=\bigcup_{i=1}^{\infty} X_i$$ where $$X_i⊂X_{i+1}⊂..., X_i$$ are open for all i and μ(X_i)<+∞ for all i. Show that for every Borel set B there exists an open set U where μ(B\U)<ϵ.

Homework Equations

measures are subadditive

The Attempt at a Solution

Pick closed sets $$c_i⊂X_i−B$$ with $$μ((X_i−c_i)−B)=μ((X_i−B)−c_i)<ϵ/{2^i}$$. Let $$G=\bigcup_i^\infty X_i-c_i$$. Then recognizing that $$X_i−c_i=X_i⋂c_i^c$$ is the intersection of open sets reveals that G must be open. Moreover, B⊂G is clear and thus we can consider G−B. If follows that μ(G−B)<ϵ by sub additivity.

My problem is explaining why such closed sets exist. I know that $$X_i-B$$ is measurable, but why am I able to find a closed set inside of it that satisfies these requirements?

 

[mighty2000]

To explain why such closed sets exist, we can use the fact that X_i is open for all i. This means that X_i^c, the complement of X_i, is closed. Since X_i is a subset of X_{i+1}, we can also say that X_{i+1}^c is a subset of X_i^c. Continuing this pattern, we can see that X_i^c is a decreasing sequence of closed sets.

Now, since μ(X_i)<+∞ for all i, we can apply the continuity property of measures to see that μ(X_i^c)→0 as i→∞. This means that for any given ϵ>0, we can choose a large enough i such that μ(X_i^c)<ϵ. This implies that there exists a closed set c_i⊂X_i^c with μ(c_i)<ϵ.

Finally, since X_i^c is a subset of X_i-B, we can say that c_i is a subset of (X_i-B)^c, which is equivalent to X_i-c_i. This shows that we can find a closed set c_i inside of X_i-B that satisfies the requirements of the problem.