Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $(X,Y)$ be random vector with probability density function $g_{(X,Y)}$. Finally, let $f$ be any borel function, such that $\mathbb E[|f(X,Y)|] < \infty$. Then it holds: $\mathbb E[f(X,Y)|Y] = h(Y)$, where:
$$ h(y) = \frac{\int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dx}{\int_{\mathbb R} g_{(X,Y)}(x,y)dx} $$when $\int_{\mathbb R} g_{(X,Y)}(x,y)dx \neq 0$, and $h(y) = 0$ otherwise.
Firstly, we can put $0$ in the second case, because the set $S=\{ \omega \in \Omega : \int_{\mathbb R} g_{(X,Y)}(x,Y(\omega)) dx = 0 \}$ has measure $0$. Clearly $\mathbb P(S) = \mathbb P(Y \in S_Y)$, where $S_Y = \{ y \in \mathbb R: \int_{\mathbb R} g_{(X,Y)}(x,y)dx = 0 \}$. Then $\mathbb P(Y \in S_Y) = \int_{S_Y} g_Y(y) dy $, where $g_Y$ is marginal density (It exists due to Fubini + existence of joint density of rv $(X,Y)$ ). But note $g_Y(y) = \int_{\mathbb R} g_{(X,Y)}(x,y)dx $, so we're just integrating $0$ function (cause we on $S_Y$ where it's $0$), so $\mathbb P(S) = 0$. This + the fact that Conditional Expectation is up to the set of measure $0$ allows us to forget about the case when $g_Y(y) = 0$.
So, we have to prove $2$ things:
1) $h(Y)$ is $\sigma(Y)$ measurable. Clearly both $\int_{\mathbb R} g_{(X,Y)}(x,Y) dx$ and $\int_{\mathbb R} g_{(X,Y)}(x,Y) f(x,Y) dx$ are $\sigma(Y)$ measurable due to Fubinii theorem (integrals of $\sigma(Y) -$ measurable functions are $\sigma(Y)$ measurable (We here used the fact that $g_{(X,Y)}$ is bounded and $\mathbb E[f(X,Y)]$ is finite to be able to apply Fubini's theorem.
2) For any $A \in \sigma(Y)$ we have to show $\int_A f(X,Y) d\mathbb P = \int_A h(Y) d\mathbb P$. Note that $A$ is of the form $Y^{-1}(B)$ where $B \in \mathcal B(\mathbb R)$ (borel set).
Note that $\int_A f(X,Y) d\mathbb P = \mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ]$ and $\int_A h(Y) d\mathbb P = \mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ]$
We'll use the fact, that if random variable/vector (in $\mathbb R^n$) $V$ has density function $g_V$, then for any borel function $\phi: \mathbb R^n \to \mathbb R^n$, we have $\mathbb E[\phi(V)] = \int_{\mathbb R^n} \phi(v) g_V(v) d\lambda_n(v)$.
Then: $$\mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ] = \int_{\mathbb R^2} f(x,y)\chi_{_{B}} g_{(X,Y)}(x,y) d\lambda_2(x,y) = \int_{B} \int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dxdy $$
That last split of integrals due to fubinii (function is integrable due to our assumption with $f$ ).
And now similarly at the beggining:
$$ \mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ] = \int_{B} h(y) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx)dy$$
Now due to our assumption of $h$ (that is getting rid of that case when denominator is $0$ due to its being $0$-measurable set). We have:
$$ \int_{B} (h(y)) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx) dy = \int_{B} (\frac{\int_{\mathbb R} g_{(X,Y)}(x,y)f(x,y)dx}{\int_{\mathbb R} g_{(X,Y)}(x,y)dx}) (\int_{\mathbb R} g_{(X,Y)}(x,y)dx )dy$$
After simplification we get $\mathbb E[ h(Y) \cdot \chi_{_{Y \in B}} ] = \int_{B} \int_{\mathbb R} f(x,y)g_{(X,Y)}(x,y)dxdy = \mathbb E[ f(X,Y) \cdot \chi_{_{Y \in B}} ]$, what we wanted to prove.
Now your "definition $1$" follows when you take $f(x,y) = x$. Then $ h(y) = \mathbb E[X|Y=y] $
Best Answer
Yes it is possible. Suppose $\nu : \Omega \times \mathcal{E} \to [0, 1]$ is a regular conditional distribution of $X$ given $\mathcal{G}$. This means that
Condition 3 says that for any $A \in \mathcal{E}$, $$E(1_A(X) \mid \mathcal{G})(\omega) = \int_{E}1_{A}(x)\nu(\omega, dx) \hspace{20pt} \text{ for a.e. } \omega \in \Omega.$$ By nonnegative linearity of conditional expectations and integrals we get that for any simple function $f : (E, \mathcal{E}) \to [0, \infty)$, $$E(f(X) \mid \mathcal{G})(\omega) = \int_{E}f(x)\nu(\omega, dx) \hspace{20pt} \text{ for a.e. } \omega \in \Omega.$$ Then for a measurable $f : E \to [0, \infty)$ with $E(f(X)) < \infty$, take a sequence of simple functions $f_n \to f$ and apply the previous result and the monotone convergence theorem for conditional expectations and integration to deduce the result for this $f$. Then apply linearity to deduce the result for real valued $f$ with $E(|f(X)|) < \infty$.