Generalised version of le Cam's Third Lemma

Question

I'm confused with the generalised version of Le Cam's Third lemma presented in Theorem 6.6 of van der Vaart asynptotics Statistics

here: https://books.google.co.uk/books?id=Ocg2AAAAQBAJ&pg=PT188&lpg=PT188&dq=van+der+vaart+theorem+6.6&source=bl&ots=Rm4Rw_KJ8H&sig=iXbKRK_HhSX1qQhKjhlGOcLC7Nc&hl=it&sa=X&ved=0ahUKEwjr1uCo2d7JAhUEJR4KHcx7CpoQ6AEIOTAE#v=onepage&q=van%20der%20vaart%20theorem%206.6&f=false

What does it mean saying that $L(B):=E(1_B(X)V)$ defines a probability measure? And how does the monotone convergence theorem help to show it?

Answer 1

Saying that $L(B):=E\left(1_B(X)V\right)$ defines a probability measure means that $L(\emptyset)=0$ and that if $\left(B_n\right)_{n\geqslant 1}$ is a sequence of measurable pairwise disjoint sets, then $$\tag{*}L\left(\bigcup_{n\geqslant 1}B_n\right)=\sum_{n=1}^{+\infty}L\left(B_n\right).$$ When the considered family is finite, this follows from linearity of expectation. For the general case, define $f_n:=\sum_{j=1}^n\mathbf 1_{B_j}(X)\cdot V$; in this way, $(f_n)_{n\geqslant 1}$ is a non-decreasing sequence and (*) follows from the monotone convergence theorem.