expected value of deterministic function

Question

I am intuitively convinced that $E[f(X)] = f(X)$, if $f$ is deterministic, but I cannot derive it using the integral definition of expectation. Assuming $X \sim unif(0,1)$ I get: $$ E[f(X)] = \int^{1}_{0}{f(x)p(x)dx} = \int^{1}_{0}{f(x)dx} = \overline{f} $$ which is not necessarily $f$.

Appreciate any help!

Edit: Thank you all, I made a mess with the notations, and somehow my question is not a question anymore

Answer 1

If $X$ is a random variable, so is $f(X)$. Unless $f$ is a degenerate function (i.e. $f(x) = const$), it is not meaningful to compare $\mathbb{E}[f(X)]$ (which is simply a number, if it exists) to $f(X)$, which is a random variable.

Answer 2

Here your pdf for the deterministic signal is an impulse of infinite amplitude and unit area..then you will get that result.

When you assume a pdf for an RV(taking infinite values) the density function is spread asymptotically..if it is deterministic then variance has to zero which intuitively suggests that it has to be an impulse