what does it mean equal in distribution Consider a discrete countable probability space $V$ and consider two probability measures such that, for any finite set
$A \subset  V$,
$$
P_1(A) = P_2 (A).
$$

How to prove that, for any bounded measurable function $f : V^{\mathbb{N}} \rightarrow \mathbb{R}$ ?
$$
E_1(f(x)) = E_2(f(x)),
$$
where $E_1( \, \cdot \,)$ is the expectation with respect to $P_1$ and 
$E_2( \, \cdot \,)$ is the expectation with respect to $P_2$? Is the boundedness assumption necessary just to ensure that the expectations are well defined?
We know that $E_1(f(x)) = \sum_{x \in V^{\mathbb{N}}}f(x) \cdot P_1(x)$ and $E_2(f(x)) = \sum_{x \in V^{\mathbb{N}}}f(x) \cdot P_2(x)$

$P_1(A) = P_2(A)$ for $A \subset V$, then we have $P_1(x) = P_2(x)$ for $x \subset V^{\mathbb{N}}$.

So the two expectations are equal.

I believe that the boundedness is necessary. If $f(x)$ is infinite for some $x$, then the corresponding term in $E_1$ or $E_2$ would be infinite if the $x$ has non-zero probability (since the distribution is discrete), and thus the expectation would not be well defined.
We know that $E_1(f(x)) = \sum_{x \in V^{\mathbb{N}}}f(x) \cdot P_1(x)$ and $E_2(f(x)) = \sum_{x \in V^{\mathbb{N}}}f(x) \cdot P_2(x)$

$P_1(A) = P_2(A)$ for $A \subset V$, then we have $P_1(x) = P_2(x)$ for $x \subset V^{\mathbb{N}}$.

So the two expectations are equal.

I believe that the boundedness is necessary. If $f(x)$ is infinite for some $x$, then the corresponding term in $E_1$ or $E_2$ would be infinite if the $x$ has non-zero probability (since the distribution is discrete), and thus the expectation would not be well defined.