the if direction：

https://math.stackexchange.com/questions/2494959/x-y-almost-surely-iff-forall-a-in-mathcalg-int-axdp-int-aydp?r=SearchResults&s=4|45.8271 

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设x，y是概率空间（Ω，F，P）上的拟可积随机变量，证明：X=Y a.e 当且仅当 xdp = ydp 对每个A∈F成立

Let (Ω,F,P)(Ω,F,P) be a probability space with G⊂FG⊂F. Let X,YX,Y be GG-measurable, and integrable. Then, how does one prove that

X=YX=Y almost surely iff ∀A∈G∫AXdP=∫AYdP∀A∈G∫AXdP=∫AYdP?

Here's my try: ∫AXdP=∫AYdP⇔∫AX−YdP=∫A(X−Y)1[X≥Y]+(Y−X)1[X<Y]dP=0∫AXdP=∫AYdP⇔∫AX−YdP=∫A(X−Y)1[X≥Y]+(Y−X)1[X<Y]dP=0

For A=[Y≥X]A=[Y≥X], we get ∫[X≥Y]X−YdP=0∫[X≥Y]X−YdP=0 which implies, by nonnegativity of X−YX−Y on A, P(X=Y)=P(X≥Y)P(X=Y)=P(X≥Y) or P(X≥Y)=0P(X≥Y)=0

For A=[Y<X]A=[Y<X], we get ∫[X<Y]Y−XdP=0∫[X<Y]Y−XdP=0 which implies, by nonnegativity of X−YX−Y on A, P(X=Y)=P(X<Y)P(X=Y)=P(X<Y) or P(X<Y)=0P(X<Y)=0

So, we get

which gives P(X=Y)=1P(X=Y)=1.

I can choose A as above since the sum of measurable functions is also measurable [X>Y]=[X−Y>0][X>Y]=[X−Y>0]

Is this a proper proof?

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  @Math1000 I've corrected the typos. – An old man in the sea. Oct 30 '17 at 11:49