Real analysis: Problem similar to uniformly integrable

[happysauce]

Homework Statement

Assume \mu(X) >0 and that f is a measurable function that maps X into ℝ and satisfies f(x) >0 for all x\inX.

Let \alpha be any fixed real number satisfying 0<\alpha<\mu(X) <infinity Prove that

inf { \int_{E}f d\mu : E\inM, \mu(E) ≥\alpha} >0.

(Hint. First prove for any \delta satisfying 0<\delta<\alpha, prov there exists n \in N such that B_{n} = {x:f(x)\geq1/n} satisfies \mu(B_{n}) \geq
\mu(X) - \delta. Then prove that if \mu(E) ≥\alpha then \mu(E\capB_{n}) ≥\alpha-\delta

Homework Equations

Not sure what is relevant.

The Attempt at a Solution

So, the hint suggests the first thing I do is show that given 0<\delta<\alpha i can find an n so that \mu(B_{n}) \geq \mu(X) - \delta. So since f(x) >0 then \mu(X) = \mu({x:f(x)>0}) = \mu({x:f(x)≥1/n} \cup{x:f(x)<1/n}). These are disjoint so we have \mu({x:f(x)≥1/n}) + \mu({x:f(x)<1/n}). If i show the delta relation with \mu({x:f(x)<1/n}) then i get \mu(B_{n}) \geq \mu(X) - \delta. It seems really trivial though and I am not sure what guarantees this.

The next part asks you to show that \mu(E\capB_{n}) ≥\alpha-\delta. Ok so \mu(E\capB_{n}) = \mu(E) +\mu(B_{n}) - \mu(E\cupB_{n})

The following hold: \mu(E\cupB_{n}) ≤\mu(X) (both are subsets)
\mu(B_{n}) ≥ \mu(X) - \delta (previous part)
\mu(E)≥\alpha (hypothesis)

So \mu(E\capB_{n}) = \mu(E) +\mu(B_{n}) - \mu(E\cupB_{n}) ≥ \mu(X) - \delta + \mu(E) - \mu(X) =\mu(E)- \delta≥\alpha-\delta


So I've done most of what the hint wants... the problem is I don't know how this helps me.

 

[happysauce]

The title is misleading, I thought it related to uniformly intagrable but I don't think it does.

 

[happysauce]

I figured out why there is an n for every δ. So I figured out both parts of the hint, but I still have no clue how to make these hints help me here. Do I have to show that

inf {\int_{B_{n}\cap E}f d\mu, E\inM, \mu (B_{n}\cap E)≥α-δ} > 0 for each n? And then as n-> infinity we have the desired result? I'm still not sure how to prove this case though...