Conditional expectation of random variable parameterized by random variable Suppose $M(\lambda)$ is a distribution that satisfies for all $Z\sim M(\lambda)$ that $E[Z]=\lambda$. Does it follow that if $X \sim M(Y)$ then $E[X | Y] = Y$ a.s.? Intuitively this seems to be the case, but how does one prove it, if it is true?
Since this has been cause of a lot of confusion; some clarification:
Fix a probability space $(\Omega,\mathcal F, P)$ so that it can accommodate the following random variables. There is only this probability space involved.
$Y$ is a random variable with expectation that takes values in $\mathbb R$ and for every $\lambda\in\mathbb R$, $M(\lambda)$ is a distribution on $\mathbb R$.
Suppose $Q_\lambda$ is the measure with distribution $M(\lambda)$. I am assuming $Q_\lambda$ to be defined on the Borel sets of $\mathbb R$. I am defining $X\sim M(Y)$ to mean that $E[1_{\{X\in A\}}|Y] = P(X\in A | Y) = Q_Y(A)$ for all measurable $A\subseteq\mathbb R$ $P$-a.s.
The conditional expectation $E[X | Y]$ is then defined as usual, as the a.s. uniquely determined random variable $H$ satisfying $E[ZH] = E[ZX]$ for all $\sigma(Y)$-measurable bounded $Z$.
 A: You say that for each $\lambda\in\mathbb{R}$, $M(\lambda)$ is a distribution on $\mathbb{R}$, by which I assume you mean a Borel probability measure. Let $\mathcal{R}$ denote the Borel $\sigma$-algebra on $\mathbb{R}$. Let us further suppose that for every $A\in\mathcal{R}$, the function $M(\cdot,A)$ is Borel measurable. Then $M$ is what's called a probability kernel from $\mathbb{R}$ to $\mathbb{R}$. It follows that $M(Y)$ is a random measure, meaning $M(Y(\omega))$ is a measure for each $\omega$ and $M(Y(\cdot), A)$ is a random variable for each $A$. It can be show that this implies $M(Y)$ is a measurable function from $\Omega$ to $M_1(\mathbb{R})$, the space of Borel probability measures on $\mathbb{R}$, where we equip $M_1$ with the $\sigma$-algebra generated by the projection maps, $\sigma(\{\pi_A: A \in \mathcal{R}\})$. (Here, $\pi_A$ is the map $\nu\mapsto \nu(A)$.)
A regular conditional distribution for $X$ given $\mathcal{G}$ is a random measure $\mu$ such that $P(X\in A\mid \mathcal{G})=\mu(\cdot,A)$ a.s. for each $A$. Your hypothesis assert that $M(Y)$ is a regular conditional distribution for $X$ given $Y$ (strictly speaking, given $\sigma(Y)$).
A special case of Theorem 5.4 in Foundations of Modern Probability by Olav Kallenberg states that if $\mu$ is a regular conditional distribution for $X$ given $Y$ and $f:\mathbb{R}^2\to\mathbb{R}$ is measurable with $E|f(X,Y)|<\infty$, then
$$
E[f(X,Y) \mid Y](\omega) = \int_\mathbb{R} f(x,Y)\,\mu(\omega,dx)
$$
for $P$-almost every $\omega\in\Omega$. This can be proven as follows. If $f=1_{A\times B}$, it follows easily from the hypotheses. Use the $\pi$-$\lambda$ theorem to prove it for all indicators $f=1_C$. Use linearity of conditional expectation to prove it for all simple $f$. Use the monotone convergence theorem for conditional expectations to prove it for all nonnegative $f$. Then apply this to the positive and negative parts of $f$.
In your case, since $\int x\,M(y,dx)=y$ for all $y$, the above yields
$$
E[X \mid Y] = \int_\mathbb{R} x \, M(Y,dx) = Y \text{ a.s.,}
$$
which is what you were hoping to conclude.
A: EDIT: See below for a response to your clarification.
Also, the notation in my original answer doesn't match the notation of your clarification, but if I try to reconcile them I'll probably introduce a typo.  So I'm leaving the original answer as is.
SECOND EDIT: We've been talking past each other this whole time!  I think I've figured it out, though.

I think there is some confusion here, so first I'm going to restate your question.
Let $(\Omega, \mathcal{F}, \mu)$ be a probability space.  Suppose that $M(\lambda)$ is a family of distributions on $\Omega$ parametrized by $\lambda \in \Lambda$.  For example, perhaps $\Omega = \mathbf{R}$, $\lambda \in \mathbf{R}$, and $M(\lambda)$ is $N(\lambda, 1)$, the unit variance normal distribution centered on $\lambda$; or perhaps $\Omega = \mathbf{R}_+$, $\lambda \in \mathbf{R}_+$, and $M(\lambda)$ is an exponential distribution with scale $\lambda$; or perhaps $\Omega = \mathbf{N}$, $\lambda \in \mathbf{N}$, and $M(\lambda)$ is Poisson with rate $\lambda$; and so on. The hypothesis made in the question is that the mean of an $M(\lambda)$-distributed random variable is $\lambda$. That is, $\Lambda \subseteq \Omega$, and if $Z \sim M(\lambda)$, then $\mathbf{E}[Z] = \lambda$ (briefly, the family is indexed by the distribution's mean).
We will need $\Lambda$ to be a probability space.  For this, we will assume $\Lambda = \Omega$ and that $\Lambda$ is endowed with same $\sigma$-algebra $\mathcal{F}$ as $\Omega$ and same measure $\mu$.
Suppose that $(\Theta, \mathcal{G}, \nu)$ is a probability space and that $Y$ is a random variable on $\Theta$ with values in $\Lambda = \Omega$.  (We need the $\sigma$-algebra on $\Lambda$ to make sense of this.)  The value of $Y$ determines a distribution $M(Y)$.  This is close to saying that $M(Y)$ is a random distribution.  However, a technical point intervenes here: For it to be a a random distribution, $M$ needs to be measurable.  By endowing $\mathfrak{M}_1(\Omega)$, the usual space of probability measures on $\Omega$, with a $\sigma$-algebra, we can make sense of this assumption.  (This is necessary to forbid pathological situations.  If $\Omega = \mathbf{R}$, $S \subseteq \mathbf{R}$ is a non-measurable set, $M(\lambda)$ is $N(\lambda, 1)$ if $\lambda \in S$, and $M(\lambda)$ is $N(\lambda, 2)$ if $\lambda \not\in S$, then $X$ isn't measurable.)  Then $M(Y)$, which means $M \circ Y$, is a random probability measure.  The event space of $M(Y)$ is $\Theta$ and the values are in $\mathfrak{M}_1(\Omega)$.
The question is: Suppose that $X = M(Y)$.  Is $\mathbf{E}[X|Y] = Y$ almost surely?  Intuitively, the mean of $X$ is the value of $Y$, so this perhaps a reasonable-looking formula.
The answer is no, because if you look at the formula a little closer it turns out to be entirely unreasonable.  We have random variables $Y \colon \Theta \to \Omega$ and $M \colon \Omega \to \mathfrak{M}_1(\Omega)$.  By definition, $X = M \circ Y \colon \Theta \to \mathfrak{M}_1(\Omega)$.  Since $X$ takes values in $\mathfrak{M}_1(\Omega)$, $\mathbf{E}[X|Y]$ must be a random probability measure on $\Omega$.  It can't equal $Y$ because $Y$ is a random element of $\Omega$.  In fact, because $X = M \circ Y$ is $\sigma(Y)$-measurable we have $\mathbf{E}[X|Y] = \mathbf{E}[X|\sigma(Y)] = X$.
(If you want to be careful about this, then you need to define $\mathbf{E}[X|Y]$; for this, recall that $\mathfrak{M}_1(\Omega)$ is contained in the topological vector space of signed measures on $\Omega$, and a topological vector space is a reasonable place in which to take expectations.)
So as stated, the question is asking about the equality of two objects of different types and is categorically false.  However, there is a way of repairing this: We simply take a different expectation.  For clarity we write expectation on $\Omega$ as $\mathbf{E}_\Omega$ and expectation on $\Theta$ as $\mathbf{E}_\Theta$.  Remember we assumed that if $Z \sim M(\lambda)$, then $\mathbf{E}_\Omega[Z] = \lambda$.  We may rewrite this as $\mathbf{E}_\Omega \circ M = 1_\Omega$, where $1_\Omega$ is the identity function on $\Omega$.  It is then trivially true that $\mathbf{E}_\Omega \circ M \circ Y = Y$.  Consequently, $\mathbf{E}_\Omega \circ X = Y$, or in more conventional notation, $$\mathbf{E}_\Omega[X] = Y.$$ It's very tempting to write this as $\mathbf{E}[X] = Y$, but $\mathbf{E}$ runs the risk of being interpreted as $\mathbf{E}_\Theta$, and as discussed above, that's a type error.  To recover a result a little more like the one you're interested in, you can apply $\mathbf{E}_\Theta[\cdot|Y]$ to both sides to get
$$\mathbf{E}_\Theta[\mathbf{E}_\Omega[X]|Y] = Y.$$

I believe my original answer is correct.  Here is what I make of your clarification.
You have, in the notation of my original answer, $\Lambda = \Omega = \mathbf{R}$; your $\Omega$ was my $\Theta$.  I'll use your notation from here on, though.  For each $\lambda \in \mathbf{R}$, there is a measure $Q_\lambda$, and $M(\lambda) = Q_\lambda$.
You define $X \sim M(Y)$ to mean that $X$ is a random variable such that $\mathbf{P}[X \in A|Y]$.  Fix $\lambda \in \mathbf{R}$.  For the sake of argument, suppose it's valid to condition on the event $Y = \lambda$ and that probability zero events don't happen.  In that case, the only thing $\mathbf{P}[X \in A|Y = \lambda]$ could be is $Q_\lambda(A)$, and that means $\mathbf{E}[X|Y = \lambda]$ is a random variable with distribution $Q_\lambda$.  In particular, it sounds like the event space of this random variable is $\mathbf{R}$.  It follows that the event space of $X$ must be $\Omega \times \mathbf{R}$: In order to actually sample from $X$, first you sample an $\omega \in \Omega$, and second you use $Q_{Y(\omega)}$ to sample from $\mathbf{R}$.
Sampling $\omega \in \Omega$ determines $Q_{Y(\omega)} \in \mathfrak{M}_1(\mathbf{R})$.  It follows that $Q_Y$ is a random measure on $\mathbf{R}$.  It's simply the composite $\omega \mapsto Y(\omega) \mapsto Q_{Y(\omega)}$.  The measure $Q_{Y(\omega)}$ is the law of a random variable, and in my original answer I conflated the random variable with its law.  It's this identification that allows me to write $X = M \circ Y$, as follows.
It's true that you wrote $X \sim M(Y)$, not $X = M(Y)$.  But $X \sim M(Y)$ means, "$X$ is a random variable with distribution $M(Y)$."  Since $X$ and $M(Y)$ have the same event space, as functions from the event space to the codomain, $X$ is a.s. equal to $M(Y)$.  Since $M(Y)$ means, "The result of applying $M$ to the value of $Y$," and since $Y$ is a measurable function $\Omega \to \mathbf{R}$, $M(Y)$ must be the measurable function $\omega \mapsto Y(\omega) \mapsto M(Y(\omega)) = Q_{Y(\omega)}$, that is, $M \circ Y$; so we conclude that $X = M \circ Y$ a.s.

I think I understand the problem now: What do we mean by $X \sim M(Y)$?  I interpreted it one way, and you interpreted it another.
In my interpretation, we sample from $M(Y)$ as follows.  We sample $\omega \in \Omega$; this determines $\lambda = Y(\omega)$; and $M(Y)$ means $M(Y(\omega)) = M(\lambda)$, which is a distribution.  To say $X \sim M(Y)$ means that the random variable $X$ is the result of applying $M$ to the random variable $Y$, and hence $X$ is a random measure.
In your interpretation, we sample from $M(Y)$ by sampling $\omega \in \Omega$; this determines $\lambda = Y(\omega)$; then we sample from $M(Y(\omega)) = M(\lambda)$.  To say $X \sim M(Y)$ means that the distribution of $X$ is the distribution $M(Y)$ determined by $Y$, and hence $X$ is a number.
Just for clarity, suppose that instead of $M$, we had a function $f \colon \mathbf{R} \to \mathbf{R}$.  In my interpretation, $X \sim f(Y)$ means that the distribution of $X$ is the result of sampling $Y$ and applying $Y$; in yours, $X \sim f(Y)$ means that we first apply $f$ to the distribution of $Y$, and the distribution of $X$ is a sample from the result.  For example, suppose $f(x) = x^2$.  I was reading $X \sim Y^2$ as, "to sample $X$, take a sample from $Y$ and square it."  You were reading it as, "there is a distribution of $Y$'s, and it determines a distribution of squares; to sample $X$, take a sample from that distribution of squares."  For real-valued $f$, it's the same either way.  But $M$ is measure-valued, and then there's a difference.
I'm not a professional probabilist or statistician, so maybe my interpretation is unorthodox.  I think that for your interpretation, I would rather write something like, "$Y \sim P$, $F = M(Y)$, and $X \sim F$."  But words are probably clearer.
Having settled that, I can answer the question you asked.  I'll be a little pedantic, just to avoid more confusion.  I hope you don't mind.
For every $\lambda \in \mathbf{R}$, we have a measure $Q_\lambda$.  We define $X_\lambda$ to be a random variable with this law.  A priori, it may have some potentially strange event space, but after pushing forward, we may replace the event space with $\mathbf{R}$.  Then $X_\lambda$ is the function from the probability space $(\mathbf{R}, Q_\lambda)$ to $\mathbf{R}$ defined by $\lambda \mapsto \lambda$.  The problem statement assumes that $\mathbf{E}[X_\lambda] = \lambda$.
We would like to construct an $X$ for which $X \sim M(Y)$.  We want a $M(Y)$-distributed sample to be realized by first sampling a value $Y(\omega)$, then sampling a value from the random variable $X_{Y(\omega)}$.  Whatever the event space of $X$ is, it must have enough room for us to specify an $\omega \in \Omega$, and once that's specified, it must have enough room for the event space of $X_{Y(\omega)}$.  This could happen, but a priori, there is no reason why it must happen.  All we know is that $\Omega$ is large enough to support the events we need for $Y$.  It might happen that $\Omega = \mathbf{R}$ and that $Y$ is the identity function; in that case, once you have specified $Y(\omega)$, you have used up all your randomness.  If we tried to use this $\Omega$ as the event space of $X$, then every $X_{Y(\omega)}$ would be deterministic, and that's clearly not what we want.
We can work around this by enlarging the event space.  I'll take the event space of $X$ to be $\Omega \times \mathbf{R}$.  I'll endow this space with the product $\sigma$-algebra, which I'll denote $\mathcal{G}$.  I'll give it a measure $\mu$ which disintegrates as $P$ on $\Omega$ and as $Q_\lambda$ on $\{\lambda\} \times \mathbf{R}$.  $X$ itself will be the function $\Omega \times \mathbf{R} \to \mathbf{R}$ given by $(\omega, \lambda) \mapsto \lambda$.  And while $Y$ is given to us as a random variable with event space $\Omega$, we may pull it back to $\tilde Y$ on $\Omega \times \mathbf{R}$ by simply projecting away the $\mathbf{R}$ coordinate: $\tilde Y(\omega, \lambda) = Y(\omega)$.  The $\sigma$-algebra $\mathcal{F}$ on $\Omega$ pulls back to $\tilde{\mathcal{F}}$ on $\Omega \times \mathbf{R}$, and $\sigma(\tilde Y) \subseteq \tilde{\mathcal{F}} \subseteq \mathcal{G}$, where $\sigma(\tilde Y)$ is the $\sigma$-algebra generated by $\tilde Y$.
Actually, after typing this all out, I notice that your clarification says that $\Omega$ "can accommodate all the following random variables."  So I guess you knew this already.  Sorry.
(And just to connect with what I wrote before: It's the disintegration of $\mu$ that defines the random measures I was discussing.)
Now we can make sense of $\mathbf{E}[X|\tilde Y]$.  (The way I've defined the event spaces, $\mathbf{E}[X|Y]$ does not make literal sense because $X$ and $Y$ are defined on different event spaces, but it's clear that $\tilde Y$ and $Y$ are effectively equivalent.)  By definition, $\mathbf{E}[X|\tilde Y]$ is $\mathbf{E}[X|\sigma(\tilde Y)]$, and by definition, $\mathbf{E}[X|\sigma(\tilde Y)]$ is a $\sigma(\tilde Y)$-measurable function $\Omega \times \mathbf{R} \to \mathbf{R}$ such that $\int_A \mathbf{E}[X|\sigma(\tilde Y)] d\mu = \int_A X\,d\mu$ for every $A \in \sigma(\tilde Y)$.
One way to get information about this conditional expectation is to use the tower property to see that $\mathbf{E}[X|\sigma(\tilde Y)] = \mathbf{E}[\mathbf{E}[X|\tilde{\mathcal{F}}]|\sigma(\tilde Y)]$.  So let's consider $\mathbf{E}[X|\tilde{\mathcal{F}}]$.  If $A \in \tilde{\mathcal{F}}$, then $A$ is of the form $B \times \mathbf{R}$ for some $B \in \mathcal{F}$.  Consequently, $$\int_A X\,d\mu = \int_B \left(\int_{\mathbf{R}} X(\omega, \lambda)\,dQ_{Y(\omega)}(\lambda)\right)\,dP(\omega).$$
From the definition of $X$, this equals $$\int_B\left(\int_{\mathbf{R}} X_{Y(\omega)}(\lambda)\,dQ_{Y(\omega)}(\lambda)\right)\,dP(\omega) = \int_B \mathbf{E}[X_{Y(\omega)}]\,dP(\omega).$$
Finally, we can apply the assumption that $\mathbf{E}(X_\lambda) = \lambda$ to get $$\int_B Y(\omega)\,dP(\omega) = \mathbf{E}_P[Y \cdot 1_B],$$
where $1_B$ is the indicator function of $B$.
From this, we may deduce that $\tilde Y$ is a representative of $\mathbf{E}[X|\tilde{\mathcal{F}}]$.  (This is analogous to when I wrote $\mathbf{E}_{\Omega}[X] = Y$ in my original post, though that was a different $\Omega$.)  To finish, we plug this into the tower property, and we get the result you wanted:
$$\mathbf{E}[X|\tilde Y] = \mathbf{E}[X|\sigma(\tilde Y)] = \mathbf{E}[\tilde Y|\sigma(\tilde Y)] = \tilde Y.$$
