Show that this sequence is positive with probability approaching one Let $(X_n)$ be a sequence of square integrable real random variables on a probability space $(\Omega,\mathcal F,P)$. Suppose that
$$E[X_n\mid \mathcal A]\to 1_A \quad P\text{-a.s.}$$
$$V[X_n\mid \mathcal A]\to 0 \quad P\text{-a.s.}$$
as $n\to \infty$, where $\mathcal A\subset \mathcal F$ is a sub $\sigma$-algebra, and $A\in\mathcal A$ has positive $P$-measure. Can I show then that
$$P[X_n\geq 0]\to 1 \quad \text{as } n\to\infty \quad?$$
Any help is very appreciated.
 A: No, I don't think we can conclude $P(X_n \ge 0) \rightarrow 1$.  Let $\Omega = [0,1]$ with $P$ the Lebesgue measure and $\mathcal A$ the sigma-algebra on $[0,1]$ generated by the Borel subsets of $[0,\frac 12]$.  Let $A = [0,\frac 12]$ and \begin{align*} X_n(x) := \begin{cases} 1 & x \in A \\
\frac 1n & x \in (\frac 12, \frac 34] \\
-\frac 1n & x \in (\frac 34,1] \end{cases}.
\end{align*}
Then $\mathbb{E}[X_n | \mathcal A] = 1_A$ for all $n$ so clearly $\mathbb{E}[X_n | \mathcal A] \rightarrow 1_A$, and $V(X_n | \mathcal A) = \frac{1}{n^2}1_{(\frac 12, 1]} \rightarrow 0$ almost surely.  However, we have $P(X_n < 0) = \frac 14$ for all $n$, so $P(X_n \ge 0) \not \rightarrow 1$.
A: In general no: let $X_n=\mathbf{1}_A+Y_n$, where $Y_n$ is independent of $\mathcal A$. The first assumption translates as $\mathbb E\left[Y_n\right]\to 0$ and the second one as $\operatorname{Var}(Y_n)\to 0$.
Moreover,
$$
\mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+Y_n\geqslant 0)+\mathbb P(A^c\cap \{Y_n\geqslant 0)
$$
and by independence,
$$
\mathbb P(X_n\geqslant 0)=\mathbb P(A)\mathbb P(1+Y_n\geqslant 0)+\mathbb P(A^c)\mathbb P(Y_n\geqslant 0).
$$
Let $(Y_n)_{n\geqslant 1}$ be a sequence of independent random variables such that for each $n$, $Y_n$ taked the values $1/n$ and $-1/n$ with probability $1/2$. Letting $\mathcal A=\sigma(Y_1)$ and $A=\{Y_1=1\}$, the assumptions on  fulfills the assumptions and for $n\geqslant 2$,
$$
\mathbb P(X_n\geqslant 0)=\mathbb P(A\cap \{1+X_n\geqslant 0\})+\mathbb P(A^c\cap \{X_n\geqslant 0\})= \mathbb P(A)+\mathbb P(A^c)/2<1,
$$
as $\mathbb P(1+Y_n\geqslant 0)=1$ and $\mathbb P(Y_n\geqslant 0)=1/2$.
