Let $X$ be a r.v. drawn from a cdf $F(x)$ with a support $[0,a]\cup [b,1]$, where $0< a \leq b < 1$. $F(x)$ is continuously twice differentiable in the following intervals: $[0,a]$ and $[b,1]$; and has a break in $(a,b),$ meaning that $X$ cannot take any values in that interval. If $b<c<1$, how do we compute the following:
$$\Pr[x \leq c]E[x|x \leq c]?$$
My attempt: \begin{align} \begin{split} \Pr[x\leq c]E[x|x\leq c] &= \Pr[x\leq a]E[x|x\leq a] + \Pr[b\leq x\leq c]E[x|b\leq x\leq c] \end{split} \end{align}
Firstly, we have $ \mathbb{P}[X \leq c] = F(c) $, by definition of the cumulative distribution function (CDF). Secondly, let us computes the expected value of $X$ under the condition $X\leq c$. To do so, we use integration by parts and the properties of the CDF: \begin{aligned} \mathbb{E}[X | X \leq c] &= \int_0^a x F'(x)\, \mathrm{d}x + \int_b^c x F'(x)\, \mathrm{d}x \, ,\\ &= a F(a) - \int_0^a F(x)\, \mathrm{d}x + cF(c) - bF(b) - \int_b^c F(x)\, \mathrm{d}x \, ,\\ &= cF(c) + (a-b)F(a) - \int_0^a F(x)\, \mathrm{d}x - \int_b^c F(x)\, \mathrm{d}x \, . \end{aligned}