If fn:X→[0,∞] is measurable, for n=1,2,3,…, and f(x)=n=1∑∞fn(x)x∈XthenX∫fdμ=n=1∑∞X∫fndμwhich also shows that ∫(f1+f2)dμ=∫f1dμ+∫f2dμ
Theorem (1.38)
Let (fn)n∈N be a sequence of C-Measurable Functions defined a.e. on X s.t. n=1∑∞X∫∣fn∣dμ<∞Then the series f(x)=n=1∑∞fn(x)converges almost for all x,f∈L1(μ) and X∫fdμ=n=1∑∞X∫fndμ
Lemma (Beppo-Levi (Probability))
For rvY1,Y2,⋯∈L1 with Yi≥0,∀i≥1, if ∑n=1∞E[Yn]=c<∞ then
\begin{proof} Let E be the event {∑n=1∞Yn=∞}. For a large constant T>0, define the rvSN=min{T,n=1∑NYn}SN is a bounded rv, SN is non-decreasing, so, it has a limit and for ω∈E, N→∞limSN(ω)=TBy Dominated Convergence Theorem we have that N→∞limE[SN]=E[N→∞limSN]=T≥T⋅P(E)But, N→∞limE[SN]≤N→∞limE[n=1∑NYn]=E[N→∞limn=1∑NYn]=cHence, P(E)≤Tc. But T is arbitrary so T→∞ yields P(E)=0.