If $A, B, C$ are events, and $P(A\vert B)=1$, prove $P(A\cap B\cap C)=P(B\cap C)$.
I have tried to prove it.
$P(A\vert B)=1\iff \dfrac{P(A\cap B)}{P(B)}=1\iff P(A\cap B)=P(B)$.
Is it right, if $$P(A\cap B)=P(B)$$ then $$A\cap B=B,$$ so $$A\cap B\cap C=B\cap C,$$ and we have $$P(A\cap B\cap C)=P(B\cap C)?$$
On
Let $X_A, X_B$, and $X_C$ be the indicator variables corresponding to the events $A, B,$ and $C$. From the given relation, we have $\mathbb{E}(X_B(1-X_A))=0$. However, due to the non-negativity of the term within expectation, we have almost surely, $X_B(1-X_A)=0$. The result now follows by multiplying both sides by $X_C$ and taking expectation.
Starting from where you got stuck: $P(A\cap B)=P(B)$ \begin{align} P(B)=P(B\setminus A)+P(A\cap B)&\implies P(B\setminus A)=0\\ &\implies 0\leq P((B\setminus A)\cap C)\leq P(B\setminus A)=0\\ &\implies P((B\setminus A)\cap C)=0\\ &\implies P(B\cap C)=P(A\cap B\cap C) +P((B\setminus A)\cap C)\\ &\implies P(B\cap C)= P(A\cap B\cap C). \end{align}