If fnā:Xā[0,ā] is measurable, for n=1,2,3,ā¦, and f(x)=n=1āāāfnā(x)xāXthenXā«āfdμ=n=1āāāXā«āfnādμwhich also shows that ā«(f1ā+f2ā)dμ=ā«f1ādμ+ā«f2ādμ
Let (fnā)nāNā be a sequence of C-Measurable Functions defined a.e. on X s.t. n=1āāāXā«āā£fnāā£dμ<āThen the series f(x)=n=1āāāfnā(x)converges almost for all x,fāL1(μ) and Xā«āfdμ=n=1āāāXā«āfnādμ
For rvY1ā,Y2ā,āÆāL1 with Yiāā„0,āiā„1, if ān=1āāE[Ynā]=c<ā then 1. n=1āāāYnā<āa.s. and consequently; 2. Ynāa.s.ā0
\begin{proof} Let E be the event {ān=1āāYnā=ā}. For a large constant T>0, define the rvSNā=min{T,n=1āNāYnā}SNā is a bounded rv, SNā is non-decreasing, so, it has a limit and for ĻāE, NāālimāSNā(Ļ)=TBy Dominated Convergence Theorem we have that NāālimāE[SNā]=E[NāālimāSNā]=Tā„Tā P(E)But, NāālimāE[SNā]āā¤NāālimāE[n=1āNāYnā]=E[Nāālimān=1āNāYnā]=cāHence, P(E)ā¤Tcā. But T is arbitrary so Tāā yields P(E)=0.