I am working on this question.
Let $X:(\Omega, \mathcal{F})\longrightarrow (\mathbb{R}, \mathcal{B})$ be a random variable with $E|X|<\infty$.
(a) Show that if $A_{n}\in\mathcal{F}$ are disjoint sets and $A:=\bigcup_{n=1}^{\infty}A_{n},$ then $\sum_{n=1}^{\infty}E[X\mathbb{1}_{A_{n}}]$ converges absolutely and $\sum_{n=1}^{\infty}E[X\mathbb{1}_{A_{n}}]=E[X\mathbb{1}_{A}].$
(b) Conclude that if $X\geq 0$, then $Q(A):=E[X\mathbb{1}_{A}]/EX$ is a probability measure.
I have solved most of it, but I don't know how to show $\sum_{n=1}^{\infty}\int_{A_{n}}X$ converges absolutely. I tried to show that this sequence is Cauchy but I failed...
Below is my proof of other parts:
(a)
Set $F_{n}:=\bigcup_{k=1}^{n}A_{k}$, then for all $n$, we have $X\mathbb{1}_{F_{n}}$ is measurable and $|X\mathbb{1}_{F_{n}}|\leq |X|$. Further, we also have $X\mathbb{1}_{F_{n}}\longrightarrow X\mathbb{1}_{A},$ as $n\rightarrow\infty.$
Thus, by Lebesgue dominated convergence theorem, we have $$\lim_{n\rightarrow\infty}\int X\mathbb{1}_{F_{n}}=\int X\mathbb{1}_{A}=\int_{A}X.$$
On the other hand, since $A_{k}$ are disjoint, we also have $$\int X\mathbb{1}_{F_{n}}=\sum_{k=1}^{n}\int X\mathbb{1}_{A_{k}}=\sum_{k=1}^{n}\int_{A_{k}}X.$$
Now, combine the above results, $$\int_{A}X=\lim_{n\rightarrow\infty}\int X\mathbb{1}_{F_{n}}=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\int_{A_{k}}X=\sum_{k=1}^{\infty}\int_{A_{k}}X.$$
This is equivalent to $$E[X\mathbb{1}_{A}]=\sum_{k=1}^{\infty}E[X\mathbb{1}_{A_{k}}].$$
(b)
Since $X$ is nonnegative, $EX\leq E|X|<\infty.$
Firstly, since $\varnothing$ is a null set, it is clear that $E[X\mathbb{1}_{\varnothing}]=0$ and thus $Q(\varnothing)=0.$ Thus, for all $A\in\mathcal{F}$, $$Q(A)=\dfrac{\int_{A}X}{\int X}\geq \dfrac{\int_{\varnothing}X}{\int X}=0.$$
Now, if $A_{k}\in\mathcal{F}$ is a countable sequence of disjoints sets such that $A:=\cup_{k=1}^{\infty}A_{k},$ then by \textbf{(a)}, $$Q(A)=E[X\mathbb{1}_{A}]/EX=\sum_{k=1}^{\infty}E[X\mathbb{1}_{A_{k}}]/EX=\sum_{k=1}^{\infty}Q(A_{k}).$$
Also, it is clear that $Q(\Omega)=E[X\mathbb{1}_{\Omega}]/EX=EX/EX=1.$
Finally, as $EX=E[X\mathbb{1}_{A}]+E[X\mathbb{1}_{A^{c}}]$, we have $$1=Q(\Omega)=E[X\mathbb{1}_{A}]/EX+E[X\mathbb{1}_{A^{c}}]/EX=Q(A)+Q(A^{c})$$, therefore we have $$Q(A^{c})=1-Q(A).$$
Therefore, $Q$ is a probability measure.
After posting my question, I suddenly figured it out... But for the reference of other users, I am gonna post my proof here.
Proof:
Firstly, we need to show that if $X_{n}\geq 0$, then $$E\Big(\sum_{n=0}^{\infty} X_{n}\Big)=\sum_{n=0}^{\infty}EX_{n}.$$
Indeed, if we define $Y_{N}:=\sum_{n=0}^{N}X_{n}$, we have $0\leq Y_{N}\nearrow Y_{\infty},$ thus by monotone convergence theorem, we have $$\lim_{N\rightarrow\infty}EY_{N}=EY_{\infty}.$$
Therefore, $$E\Big(\sum_{n=1}^{\infty}X_{n}\Big)=EY_{\infty}=\lim_{N\rightarrow\infty}EY_{N}=\lim_{N\rightarrow\infty}\sum_{n=0}^{N}EX_{n}=\sum_{n=0}^{\infty}EX_{n}.$$
Now, applying above result, we have $$\sum_{n=0}^{\infty}|E(X\mathbb{1}_{A_{n}})|\leq \sum_{n=0}^{\infty}E(|X\mathbb{1}_{A_{n}}|)=\sum_{n=0}^{\infty}E(|X|\mathbb{1}_{A_{n}})=E\Big(\sum_{n=0}^{\infty}|X|\mathbb{1}_{A_{n}}\Big)\leq E|X|<\infty.$$
Therefore, $\sum_{n=1}^{\infty}E[X\mathbb{1}_{A_{n}}]$ converges absolutely.