To compute the expectation $E[X|X \le c]$ can I say $$ = \int\limits_{0}^{c} x \ p(x|x\le c) \ dx \ + \ 0 = \int\limits_{0}^{c} x \frac{p(x,x\le c)}{p(x\le c)} dx = \int\limits_{0}^{c} x \frac{p(x)}{p(x\le c)} dx $$
In essence, is it valid to say $p(x,x \le c) = p(x) \ in \ [0,c]$ ?
Is there a way to take $X \le c$ as an indicator function on $X$ and compute the same?
Presumably, you are truncating your random variable, and you want to compute the expected value of the truncated version. The condition $x\le c$ does not create a "conditional distribution".
In general, consider a continuous r.v. $X$ with probability density function $f_X(x)$ and cumulative distribution function $F_X(x)$, and with support, say $[a,b]$ (meaning that outside this range the probability density function is defined equal to zero, and that $F_X(a) = 0, \; F_X(b) =1$).
If we want to truncate the support at say, a subinterval $[c_1, c_2] \subset [a,b]$, then the probability density function of this truncated version is
$$f_X^{(tr)}(x) = \frac{f_X(x)}{F_X(c_2) - F_X(c_1)} $$
...i.e. we "normalize" the density function in order to integrate to unity over the domain $[c_1, c_2]$.
Then, the expected value of this truncated r.v. is
$$E^{(tr)}(x) = \int_{c_1}^{c_2} x\frac{f_X(x)}{F_X(c_2) - F_X(c_1)}dx $$