I'm trying to show that, given an integrable random variable in $\Bbb R$, for all $\lambda \ge 0$, $$\Bbb{E}(e^{\lambda X} | X \le K) \uparrow \Bbb{E}(e^{\lambda X}).$$ I can show that the left hand side is bounded above by the right hand side, but I can neither show that it is increasing nor that the limit is actually the right hand side.
This is what I have so far. I just did the standard thing, writing $Y = e^{\lambda X}$: \begin{align} E(Y) &= E(Y|X \le K)P(X \le K) + E(Y|X > K)P(X > K) \\ &= E(Y|X \le K) + (E(Y|X > K) - E(Y|X \le K))P(X>K) \\ &\ge E(Y|X \le K), \end{align} since $E(Y|X \le K) \le e^{\lambda K} \le E(Y|X > K)$. Further, since $X$ is integrable, we know that $P(X > K) \to 0$ and $K \to \infty$.
This is as far as I can get. I am having issues since $E(Y|K>K)$ is unbounded as $K \to \infty$ (if $\lambda > 0$; in fact, as above, $E(Y|X>K) \ge e^{\lambda K} \to \infty$ as $K \to \infty$. I can't even see why necessarily $(E(Y|X > K) - E(Y|X \le K))P(X>K)$ is decreasing as a function of $K$.
Any advice would be most appreciated. However, if you've seen my questions before, you may know that I have a general disclaimer: please do not post full solutions! I'm trying to learn maths, and just having someone tell me the answer doesn't really help. If I just wanted the answer, I'd wait until my supervision this week. What I would like is a very minor push in the right direction; I can then (hopefully!) do the rest myself, and learn from it. Thank you.
PS: Sorry about the awful title -- if anyone can think of a better one, then please just change it and (providing it is better!) I'll accept the change.