Proving Integrability on [0,1]: A Convergence Analysis

[bbkrsen585]

This is a question I have been struggling with for some days now, but have not been able to answer.

Suppose g_n are nonnegative and integrable on [0, 1], and that g_n $$\rightarrow$$ g almost everywhere.

Further suppose that for all $$\epsilon$$ > 0, $$\exists$$ $$\delta$$ > 0 such that for all A $$\subset$$ [0, 1], we have

meas(A) < $$\delta$$ implies that sup_n $$\int$$_A |g_n| < $$\epsilon$$.

Prove that g is integrable, and that $$\int$$_[0,1] g = lim $$\int$$_[0,1] g_n.

I know that I need to find some way to bound g, but am unsure of how.

 

[lippyka]

Any help would be greatly appreciated. Answer: Since gn \rightarrow g almost everywhere and the sequence of functions (gn) is nonnegative, integrable and uniformly bounded, we have by the Lebesgue Dominated Convergence Theorem that g is integrable and that \int[0,1] g = lim \int[0,1] gn. To prove that g is integrable, we need to find an integrable function h such that |g| \leq h almost everywhere. Let \epsilon > 0 be given. By assumption, there exists \delta > 0 such that for all A \subset [0, 1], if meas(A) < \delta, then supn \intA |gn| < \epsilon. Since each gn is integrable, there exists a measurable function hn such that |gn| \leq hn almost everywhere. Define h = maxn hn. Then h is integrable and for each n, |gn| \leq h almost everywhere. Since g = limn gn almost everywhere, it follows that |g| \leq h almost everywhere, so g is integrable.