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At Resnick's book "a probability path" at p. 134. Given a random variable $X\geq 0,$ show that $\int_AXdP=0$ if and only if $P(A\cap [X>0])=0$.

I was trying the following procedures:

Assume $\int_AXdP=0$. We have

$E(X1_A)=E(X1_A1_{[X=0]})+E(X1_A1_{[X>0]})=0.$

So $E(X1_A1_{[X>0]})=E(X1_{[A\cap [X>0]]})=0$ since $E(X1_A1_{[X=0]})=0.$

Then how should I use the above result to show $E(1_{[A\cap [X>0]]})=0?$ Any advice would be appreciated.

lzstat
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1 Answers1

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$$\int_A X dP = 0 \iff P(A \cap \{X>0\})=0$$

Consider the following equalities: $\int_AXdP = \int_{A \cap \{X>0\}} XdP + \int_{A \cap \{X=0\}}XdP = \int_{A \cap \{X>0\}} XdP$

"$\Leftarrow$": you are calculating an integral on a set of measure $0$, so the integral is $0$

"$\Rightarrow$": $0 = \int_AXdP = \int_{A \cap \{X>0\}} XdP$.

Now use the lemma (the answer there works for measure spaces) "if $\mu(\Omega)>0, f>0$ then $\int_\Omega f d\mu>0$".

dcolazin
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  • Ty. Given that a r.v. $X>0$. If $P(A)>0,$ then $\int_A X dP>0.$ Using contrapositive, if $\int_A X dP=0,$ we have $P(A)=0.$ I am now able to prove the contrapositive part as follows: 1) $1_{(A\cap [X>\epsilon])}\leq X/\epsilon 1_{(A\cap [X>\epsilon])} \leq x/\epsilon 1_A $; 2) Taking expectation, we have $P{(A\cap [X>\epsilon])}\leq\epsilon^{-1} E{X1_A}=0$ for arbitrary $\epsilon>0$; 3) Letting a sequence $\epsilon_k(>0)\rightarrow 0$, k=1,2,..., then the monotone seq property implies that $P(A\cap [X>0])=lim_{n\rightarrow \infty}P(A\cap[X>\epsilon_n])=0$ by (2). It is very deep, right? – lzstat Aug 05 '19 at 21:35