I am trying to prove this claim.
Short version:
Let $\ X$ be an almost surely positive random variable (i.e. $\ X > 0$ a.s.) defined on the probability space $\ (\Omega, \mathcal G, P)$. Let $\mathcal F$ be a sub $\sigma$-algebra of $\mathcal G$, then
$\ Y = E[X|\mathcal F] > 0$ a.s.
Long version:
Let $\mathcal F(t), 0\le t \le T, $ be a filtration. Define $\ V(t) = E[V(T)\ exp{(-\int_t^T R(u)du)}\ |\mathcal F(t)]$, and assume
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$\ V(T)$ is almost surely positive,
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$\ R(t)$ is an adapted process,
We are asked to show that $\ V(t)$ is almost surely positive. If one defines $\ X = V(T)\ exp{(-\int_t^T R(u)du)} $, and given that $\ V(T)$ and $\ exp{(-\int_t^T R(u)du)}$ are almost surely positive random variables, we obtain the short version.
Best Answer
Let $F = \{E(X \mid \mathcal{F}) \leq 0 \}$. Then $F \in \mathcal{F}$ because, by definition, $E(X \mid \mathcal{F})$ is $\mathcal{F}$-measurable.
Now suppose $P(F) > 0$. Using the fact that $X > 0$ a.s. and the definition of conditional expectation we get $$0 < \int_F X dP = \int_F E(X \mid \mathcal{F}) dP \leq 0.$$ That's a contradiction, so $P(F)=0$. Taking complements we conclude that $E(X \mid \mathcal{F}) > 0$ a.s.