Is it true that $ \mathbb{E}(X\mid X\leq x)\leq \mathbb{E}(X\mid A)$ whenever $\mathbb{P}(X\leq x)\leq \mathbb{P}(A)$? A proof I am trying to understand seems to take the following for granted.

Let $X\in L^1(\Omega, \mathcal{F},\mathbb{P}$) be a continuous r.v. and $x\in\mathbb{R}$ be such that $\mathbb{P}(X\leq x)>0$. Then for all $A\in\mathcal{F}$ with $\mathbb{P}(A)\geq \mathbb{P}(X\leq x)$ we have $$\mathbb{E}(X\mid A)\geq \mathbb{E}(X\mid X\leq x).$$

I doubt it is true, and I hope you could help me clarify it.
 A: The claim is true. First note that $\mathbb{P}(A)\geq \mathbb{P}(X\leq x)$ implies $\mathbb{P}(X\leq x\mid A) \leq \mathbb{P}(A \mid X\leq x)$:
\begin{equation}
\mathbb{P}(A)\geq \mathbb{P}(X\leq x) \Rightarrow \frac{\mathbb{P}(A\cap X\leq x)}{\mathbb{P}(A)}\leq \frac{\mathbb{P}(A\cap X\leq x)}{\mathbb{P}(X\leq x)} \Rightarrow \mathbb{P}(X\leq x\mid A) \leq \mathbb{P}(A \mid X\leq x).
\end{equation}
Now, let us decompose the expectation $\mathbb{E}(X\mid A)$ in to two parts (using the tower law with the condition $X\leq x$):
\begin{equation}
\mathbb{E}(X \mid A) = \mathbb{P}(X\leq x\mid A)\mathbb{E}(X\mid X\leq x \cap A) + (1-\mathbb{P}(X\leq x \mid A))\mathbb{E}(X \mid X > x \cup A)
\end{equation}
Apply the lemma $\mathbb{P}(X\leq x\mid A) \leq \mathbb{P}(A \mid X\leq x)$ as well as $\mathbb{E}(X \mid X> x \cup A) \geq x \geq \mathbb{E}(X \mid X\leq x \cap A)$:
\begin{equation}
\geq \mathbb{P}(X\leq x\mid A)\mathbb{E}(X\mid X\leq x \cap A) + (1-\mathbb{P}(X\leq x \mid A))\mathbb{E}(X \mid X > x \cap A) - ( \mathbb{P}(A \mid X\leq x) - \mathbb{P}( X\leq x \mid A) ) (\mathbb{E}(X \mid X > x \cap A) - \mathbb{E}(X \mid X \leq x \cap A)),
\end{equation}
where the subtracted term is non-negative as it is a product of two non-negative factors. Next, simplify the expression:
\begin{equation}
= \mathbb{P}(A \mid X\leq x) \mathbb{E}(X\mid X\leq x \cap A) + (1 - \mathbb{P}(A \mid X \leq x))\mathbb{E}(X \mid X>x \cap A)
\end{equation}
Then, apply $\mathbb{E}(X \mid X>x \cap A)\geq x \geq \mathbb{E}(X \mid X\leq x \cap \neg A)$:
\begin{equation}
\geq \mathbb{P}(A \mid X\leq x) \mathbb{E}(X\mid X\leq x \cap A) + (1 - \mathbb{P}(A \mid X \leq x))\mathbb{E}(X \mid X\leq x \cap \neg A).
\end{equation}
This last expression is the decomposition of $\mathbb{E}(X \mid X\leq x)$ into two parts conditioning on $A,\neg A$. Thus, it has been shown that 
\begin{equation}
\mathbb{E}(X \mid A) \geq \mathbb{E}(X \mid X\leq x).
\end{equation}
