Suppose I have a general random X, that is not necessarily continuous. It makes sense to me that I should be able to write:
$$E(X|X>c)=\dfrac{E(X\cdot 1_{X>c})}{P(X>c)}$$
If I assume that the variable is either completely discrete or continuous, the conditional probability/density definitions can be used. But I can't find an argument for a general distribution.
The solution hinges on how $\mathbf{E}[\cdot \mid X > c]$ is defined.
If it is defined as the expectation with respect to the conditional probability measure $\mathbf{P}(\cdot \mid X > c)$, then the proof goes by invoking the standard machinery as follows:
Theorem. For any random variable $Y$ that is either a.s.-non-negative or integrable with respect to $\mathbf{P}$, we have
$$ \mathbf{E}[Y \mid X > c] = \frac{\mathbf{E}[Y \cdot \mathbf{1}_{\{X > c\}}]}{\mathbf{P}(X > c)} \tag{*}\label{e:wtp} $$
Step 1. Note that if $\mathbf{Q}$ is any probability measure and $\mathbf{E}_{\mathbf{Q}}$ is the associated expectation, then we have $\mathbf{E}_{\mathbf{Q}}[\mathbf{1}_A] = \mathbf{Q}(A) $ for any event $A$. Now applying this to the choice $\mathbf{Q}(\cdot) = \mathbf{P}(\cdot \mid X > c)$, for any event $A$ we get
\begin{align*} \mathbf{E}[\mathbf{1}_A \mid X > c] = \mathbf{P}(A \mid X > c) &= \frac{\mathbf{P}(A \cap \{ X > c \})}{\mathbf{P}(X > c)}. \end{align*}
Then applying the same argument for $\mathbf{Q} = \mathbf{P}$, the right-hand side becomes
\begin{align*} \frac{\mathbf{P}(A \cap \{ X > c \})}{\mathbf{P}(X > c)} &= \frac{\mathbf{E}[\mathbf{1}_{A \cap \{ X > c \}}]}{\mathbf{P}(X > c)} = \frac{\mathbf{E}[\mathbf{1}_A \cdot \mathbf{1}_{\{ X > c \}}]}{\mathbf{P}(X > c)}. \end{align*}
This proves that $\eqref{e:wtp}$ holds when $Y = \mathbf{1}_A$ is an indicator function.
Step 2. By the linearity of expectation, $\eqref{e:wtp}$ holds when $Y$ is a linear combination of finitely many indicator functions.
Step 3. Now let $Y$ be any non-negative random variable. Then
$$ Y_n = \min\biggl\{ \frac{\lfloor 2^n Y \rfloor}{2^n}, n \biggr\} = \sum_{k=0}^{n2^n - 1} \frac{k}{2^n} \mathbf{1}_{\{\frac{k}{2^n} \leq Y < \frac{k+1}{2^n} \}} + n \mathbf{1}_{\{Y \geq n\}} $$
increases monotonically to $Y$ as $n \to \infty$. Also, by Step 2,
\begin{align*} \mathbf{E}[ Y_n \mid X > c] &= \frac{\mathbf{E}[Y_n \cdot \mathbf{1}_{\{X > c\}}]}{\mathbf{P}(X > c)}. \end{align*}
Then by letting $n \to \infty$, the monotone convergence theorem tells that $\eqref{e:wtp}$ holds in this case.
Step 4. If $\mathbf{E}[|Y|] < \infty$, then writing $Y = Y_+ - Y_-$ for the positive part $Y_+ = \max\{Y, 0\}$ and the negative part $Y_- = \max\{-Y, 0\}$ and then applying Step 3 to each of $Y_{\pm}$ shows that $\eqref{e:wtp}$ holds in this case.
By definition of $E[X\mid X>c]$:
$$E[X\mid\sigma(X>c)]=E[X\mid X>c]1_{X>c}+E[X\mid X\leq c]1_{X\leq c}$$
And by definition of conditional expectation:
$$E[X1_{X>c}]=E\big[E[X\mid\sigma(X>c)]1_{X>c}\big]=E\big[E[X\mid X>c]1_{X>c}\big]=$$
$$=E[X\mid X>c]E[1_{X>c}]=E[X\mid X>c]P(X>c)$$
I would like to clarify this point:
Let $X$ be a random variable defined on $(\Omega,\mathcal{F},\mathbf{P})$ and $\mathcal{A},\mathcal{H}\in \mathcal{F}$
According to the definition of conditional expectation, it is easy to show that:
$$\mathbf{E}[X\mid\sigma(\mathcal{H})]=\mathbf{E}[X\mid I_{\mathcal{H}}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$
We call these constants $\mathbf{E}[X\mid\mathcal{H}]$ and $\mathbf{E}[X\mid \mathcal{\bar{H}}]$, respectively. That is to say:
$$\mathbf{E}[X\mid\mathcal{H}]=\frac{\mathbf{E}[XI_\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}}]=\frac{\mathbf{E}[XI_\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$ Similarly: $$\mathbf{P}[\mathcal{A}\mid\sigma(\mathcal{H})]=\mathbf{P}[\mathcal{A}\mid I_{\mathcal{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}I_{\mathcal{H}}+\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}I_{\mathcal{\bar{H}}}$$
We call these constants $\mathbf{P}[\mathcal{A}\mid\mathcal{H}]$ and $\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]$, respectively. That is to say:
$$\mathbf{P}[\mathcal{A}\mid\mathcal{H}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{H}]}{\mathbf{P}(\mathcal{H})}\quad and \quad\mathbf{P}[\mathcal{A}\mid \mathcal{\bar{H}}]=\frac{\mathbf{P}[\mathcal{A}\cap\mathcal{\bar{H}}]}{\mathbf{P}(\mathcal{\bar{H}})}$$
And, since: $\mathbf{E}[X\mid\sigma(\mathcal{H})]=\displaystyle\int Xd\mathbf{P}(.\mid\sigma(\mathcal{H})) $, it follows easily that: $$\mathbf{E}[X\mid\mathcal{H}]=\int Xd\mathbf{P}(.\mid\mathcal{H})\quad and \quad \mathbf{E}[X\mid \mathcal{\bar{H}}]=\int Xd\mathbf{P}(.\mid\mathcal{\bar{H}})$$
I will expand on Sangchul's comment that the solution depends on how we define $\mathbf{E}\left( X \mid X \in B \right)$, where $B$ is any Borel set. Suppose we work in a probability space $\left( \Omega, \mathcal{F}, \mathbf{P} \right)$ and let $X, Y$ be Borel-measurable random variables.
For any $\mathcal{H} \subset \mathcal{F}$ we define the conditional expectation $\mathbf{E}\left( X \mid \mathcal{H} \right)$ to be any $\mathcal{H}$-measurable random variable such that
$$\int_{H} X \ \text{d}\mathbf{P} = \int_H \mathbf{E}\left( X \mid \mathcal{F} \right) \ \text{d}\mathbf{P} \quad \forall H \in \mathcal{H}$$
We further extend this definition by understanding $\mathbf{E}\left( X \mid Y \right)$ to mean $\mathbf{E}\left( X \mid \sigma_Y \right)$.
One option is to interpret $\mathbf{E}\left( X \mid X \in B \right)$ as $\mathbf{E}\left( X \mid \mathbf{1}_{X \in B} \right)$, ie. conditioned on a random variable. By the preceding comments this makes $\mathbf{E}\left( X \mid X \in B \right)$ a well-defined random variable.
Another reasonable choice is to regard $\mathbf{E}\left( X \mid X \in B \right)$ as $\mathbf{E}\left( X \mid X^{-1}\left( B \right) \right)$, ie. conditioned on an event. We still must define what this means, and I think Sangchul gives a great answer using a conditional probability measure.
A neat property I think worth mentioning concerns consistency between the two interpretations. It can be show that for any $A \in \mathcal{F}$
$$\mathbf{E}\left( X \mid \mathbf{1}_A \right)\left( \omega \right) = \mathbf{E}\left( X \mid A \right)$$
for almost all $\omega \in A$.
