the answer for your question is yes. Note that, all you need to prove is the following equality
$$
\int_{E} X\ dP
=
\int_{E} \left[\int_{\Omega} X\ dP(\cdot|\mathcal{N})(\omega) \right]
dP|_{\mathcal{N}} \qquad \qquad (1).
$$
for all $E\in\mathcal{N}$.
The simplest way to do this is start from characteristic functions and then use linearity and finally the powerfull approximation theorems of the measure theory.
Let us start with the case $X=1_{A}$ where $A\in \mathcal{F}$. In this case the above equality holds as long as
$$
P(A\cap E)=\int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right]
dP|_{\mathcal{N}}.
$$
But this is certainly true and we can prove this manipulating the rhs. Indeed
$$
\int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right]
dP|_{\mathcal{N}}
=
\int_{\Omega} \mathbb{E}(A\cap E|\mathcal{N})
dP|_{\mathcal{N}}
=
\int_{E} \mathbb{E}(A|\mathcal{N})
dP|_{\mathcal{N}},
$$
where we used in the last equality that $E$ is $\mathcal{N}$ measurable.
But the right hand side above is by the properties of conditional expectation equals to $P(A\cap E)$. So the identity (1) holds for any characteristic function and for all $E\in\mathcal{N}$.
By linearity of the integral (1) is certainly true if $X$ is any simple function. Let $X$ be any positive $\mathcal{F}-$measurable function. We know that there is a sequence of simple functions $\varphi_n \uparrow X$, here the up arrow means monotone convergence. Since at this point is clear for us that
$$
\begin{eqnarray}
\int_{E} \varphi_n\ dP
&=&
\int_{E} \left[\int_{\Omega} \varphi_n\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}}
\\
&=& \int_{E} \mathbb{E}(\varphi_n|\mathcal{N}) dP|_{\mathcal{N}},
\end{eqnarray}
$$
we can finish the proof by invoking the Monotone Convergence Theorem.
Edition. I corrected the error pointed out by Didier Piau.
Addition. Theorem. Let $(\Omega,\mathcal{F},\mu)$ be a $\sigma$-finite measure space, $\mathcal{A}$ a sub-$\sigma$-algebra of $\mathcal{B}$, and $\nu=\mu|_{\mathcal{A}}$. If $f\in L^1(\mu)$, there exist $g\in L^1(\nu)$ (thus $g$ is $\mathcal{A}$-measurable) such that
$$
\int_{E} f\ d\mu =\int_{E} g\ d\nu
$$
for all $E\in A$; if $g'$ is another such function then $g=g'$ $\nu$-a.e. (Note that $g$ is the conditional expectation of $f$ on $\mathcal{A}$. (Folland)
This got pretty long but it's the way I think about it.
A $\sigma$-algebra represents information. Formally it's a set of events, but you can think of it as a set of questions you know the answers to.
Conditional expectation is a way of making sense of the idea that if you know some information (represented by a $\sigma$-algebra) you get a new probability distribution conditioned on that information.
So if my $\sigma$-algebra represents a list of questions I want to associate a probability distribution with every possible set of answers. Doing this rigorously presents all sorts of problems and conditional expectation is a formal tool to get around them. To lay out the basic idea I'm going to look as the finite case on more detail.
Suppose you have a probability space $(\Omega, \mathcal F, \mathbb P)$. I can generate a finite $\sigma$-algebra $\mathcal G$ from a finite set of events $(E_1, \dots, E_n)$.
We can interpret this another way, suppose I've chosen an element $\omega\in\Omega$ and you're trying to guess what it is. To make it easier I'm going to let you ask $n$ questions.
You have to choose all $n$ questions before you start asking them in this version of the game.
You've chosen the questions "is $\omega$ in $E_1$?" ... "is $\omega$ in $E_n$?".
When you've asked all your questions you have a conditional probability distribution on $\Omega$. In there are $2^n$ different sets of answers, so there are up to $2^n$ different probability distributions to consider. (some combinations of answers may occur with probability $0$).
So we could have function from the set of answer sequences to the set of probability distributions on $\omega$. But as some combinations of answers may not yield a well defined conditional probability it's better to think of this as a function from $\Omega$ to the set of probability distributions defined $\mathbb P$ almost everywhere. That is every $\omega$ give me a set of answers, so I associate with $\omega$ the probability distribution conditioned on those answers. As there are only finitely many answer sequences, with probability one I get a sequence of answers that occurs with positive probability. So I get a well defined conditional distribution with probability one.
Now let $\ell_1$ be the set of $\mathcal F$-measurable functions.
We can always define a probability distribution in terms of its expectation operator.
Which is a linear functional $\ell_1\to[-\infty,\infty]$.
So if I want to define a useful mathematical object that associates a probability distribution with (almost) every $\omega$ I can define a conditional expectation operator
$$\mathbb E(\circ |\mathcal G)(\circ): \ell_1\times\Omega\to[-\infty,\infty].$$
I can think of this in two ways, either as assigning an expectation operator to every $\omega\in\Omega$ or associating a random variable $\mathbb E(f|\mathcal G):\Omega\to[-\infty,\infty]$ with every $f\in\ell_1$.
As I've only defined my conditional distributions for almost every $\omega$ it's better to use the second idea and think of conditional expectation as a map $\ell_1\to\ell_1$, because random variables only need to be defined almost everywhere.
So to get around the almost everywhere problem we say a well defined function $\ell_1\times\Omega\to[-\infty,\infty]$ is a conditional expectation if it gives the right random variables $\mathbb E(f|\mathcal G):\Omega\to[-\infty,\infty]$
So we have a choice of different versions of conditional expectations, we need a test to check if a given operator gives the right random variables.
In this case we need to find necessary and sufficient conditions for a conditional expectation to agree with the classical case almost everywhere.
Notice three things, firstly for every $f\in\ell_1$ the conditional expectation $\mathbb(f|\mathcal G):\Omega\to[-\infty,\infty]$ must be $\mathcal G$ measurable, because $\mathbb(\circ|\mathcal G)(\omega)$ is an expectation operator associated with a conditional probability distribution which depends on the answers to the $n$ questions.
Secondly you can check that for every function $f$ we must have $\mathbb E\left(\mathbb E(f|\mathcal G)\right) = \mathbb E(f)$.
Thirdly if $g$ is a $\mathcal G$-measurable function then $\mathbb E(fg|\mathcal G)(\omega) = g(\omega)\mathbb E(f|\mathcal G)(\omega)$ because $g$ is almost surely constant on each of the (finitely many) expectation operators $\mathbb E(\circ|\mathcal G)(\omega)$.
You should be able to convince yourself that for finite $\mathcal G$ the classical definition of conditional expectation is the only function that satisfies these three conditions.
As conditional expectation is only defined almost everywhere, the second condition doesn't make sense. But as we have completely free choice of $\mathcal G$-measurable $g$ we can combine the last two conditions to get
$$\mathbb E\left( \mathbb E(fg|\mathcal g)\right) = \mathbb E\left(\mathbb E(f|\mathcal G)g\right).$$
Again convince yourself that anything satisfying this must agree with the classical conditional expectation almost everywhere.
So for a finite $\sigma$-algebra the classical conditional expectation
can be described as the only $\mathcal G$-measurable function that satisfies the condition above.
For the finite case this is all a bit unnecessary, but it works. If I define a conditional expectation operator $\mathbb E(\circ |\mathcal G)(\circ): \ell_1\times\Omega\to[-\infty,\infty]$ this will give me an expectation operator
and hence a probability distribution for almost every $\omega$.
Furthermore that conditional probability distribution will agree with
the classical one for almost every choice of $\omega$.
Suppose instead of $n$ questions I allowed you to ask a countably infinite list of questions. Now we want to do the same thing, but now the set of possible answers is uncountable. So it's quite possible that every set of answers will have probability $0$ and I can't use the normal definition of conditional expectation.
But what I want to achieve is the same thing. For every set of answers I want a conditional distribution on my probability space given those answers.
The ideas above still work with infinite sigma algebras, but you need to mess about with Radon-Nykodym derivatives to prove it and I assume you're familiar with that.
But it turns out there always exists a conditional expectation operator that satisfies the two conditions.
So, although formally we describe conditional expectation as a random variable associated with each $\mathcal F$ measurable function, anything that satisfies the conditions gives me a probability distribution for almost every $\omega$. I can interpret that distribution as the conditional distribution given that I "know" $\mathcal G$.
Best Answer
Let $(\Omega,\mathcal{F},P)$ be a probability space. Suppose $X$ is an integrable random variable and let $\mathcal{G}$ be a sub-sigma-field of $\mathcal{F}$. The conditional expection $E[X\mid\mathcal{G}]$ is the unique random variable that satisfies:
1) $E[X\mid\mathcal{G}]$ is $\mathcal{G}$-measurable.
2) $\int_A E[X\mid\mathcal{G}]\,\mathrm dP = \int_A X\,\mathrm dP$ for all $A\in\mathcal{G}$.
I'm assuming you want to find expressions for $E[X\mid\mathcal{F}]$ and $E[X\mid\{\emptyset,\Omega\}]$. For the first conditional expectation, try showing that $X$ satisfies 1) and 2), and for the last conditional expectation, try with $E[X]$.
One some times sees $E[X\mid\mathcal{G}]$ as our best guess of $X$ given the information contained in $\mathcal{G}$. Try holding this up with the expectations when $\mathcal{G}=\mathcal{F}$ and $\mathcal{G}=\{\emptyset,\Omega\}$.