How can I show that the conditional expectation $E(X\mid X)=X$? I tried to show that $E(X\mid X=x)=x$, which would lead me to get $E(X\mid X)=X$ but I am having trouble doing so. I know that the definition of conditional expectation (continuous case) is: $$E(X\mid Y=y)=\int_{-\infty}^{\infty}x f_{X\mid Y}(x\mid y)\,dx$$
This led me to write:
$$E(X\mid X=x)=\int_{-\infty}^{\infty}x f_{X\mid X}(x\mid x)\,dx$$ but I don't know what $f_{X\mid X}(x\mid x)$ is (if that's even a valid notation...)

I do note though that I don't know anything about $\sigma$-fields or Borel sets, as someone had tried to explain me once. Is it possible to prove this in an "elementary" way?
It is actually easier to define $E(X\mid Y)$ than $E(X\mid Y=y)$ hence let us recall how this is done. By definition, $E(X\mid Y)$ is any random variable $Z$ such that:

$\qquad$ (1.) $Z$ is a measurable function of $Y$.

$\qquad$ (2.) For every bounded random variable $v(Y)$, $E(Zv(Y))=E(Xv(Y))$.

It happens that when $X$ is integrable, such a random variable $Z$ exists and is almost surely unique, in the sense that if $Z$ and $Z'$ both satisfy conditions (1.) and (2.), then $P(Z=Z')=1$.

When $X=Y$, things are quite simple since $Z=X$ obviously satisfies conditions (1.) and (2.), thus, $E(X\mid X)=X$. 

Note finally that $Z=X$ always satisfies (2.) but does not satisfy (1.) in general, and, likewise, that $Z=E(X)$ always satisfies (1.) but does not satisfy (2.) in general.
There's another way to see this.

You can define $ h(x) = E(X|X=x) $ in such a way that $ E(X|X) = h(X) $. Following this logic, the expectation of $X$ given that $X$ takes on the particular value $x$ is $x$, because the expectation of a constant function is the constant function itself, i.e., $$h(x) = E(X|X=x) = x,$$ and then we conclude that $$ E(X|X) = h(X) = X. $$