If one stochastic process is a modification of another, then they have the same finite probability distribution. On page 2 in Karatzas and Shreve: Brownian Motion and Stochastic Calculus it is said that a stochastic process $Y$ is a modification of $X$ if for all $t$:
$P(X_t=Y_t)=1$.
If both are stochastic processes into $\mathbb{R}^d$ it is said that they have the same finite dimensional distribution if for every $n$ and every $A \in \mathcal{B}(\mathbb{R}^{nd})$, and any real numbers $0 \le t_1<t_2< \ldots<t_n <\infty$ we have 
$P((x_{t_1},x_{t_2},\ldots,x_{t_n})\in A)=P((y_{t_1},y_{t_2},\ldots,y_{t_n})\in A)$
Is it trivial that the first condition gives the second condition? How do we prove it? I think that in order to show this, you are not supposed to used the facts learned in elementary probability and statistics classes, but use product measures of some kind? Does it in some way follow from the theory that concerns the extensions of measures and product measures?
 A: If $(y_t)$ is a modification of $(x_t),$ then the right hand side of (1) 
equals zero:
\begin{eqnarray*}\Bigl|\mathbb{P}((x_{t_1},\ldots,x_{t_n})\in A)-\mathbb{P}((y_{t_1},\ldots,y_{t_n})\in A)\Bigr|&\leq& \mathbb{P}\Bigl((x_{t_1},\dots,x_{t_n})\neq (y_{t_1},\dots,y_{t_n})\Bigr)\\ &\leq&\sum_{i=1}^n \mathbb{P}(x_{t_i}\neq y_{t_i}).\tag1
\end{eqnarray*}
A: Perhaps the $\pi-\lambda$ theorem is more appropriate here.
First, define the $\lambda-$ system to be the set of all the 
$A \in \mathcal{B}(\mathbb{R}^{n})$, which satisfies the condition: for any real numbers $0 \le t_1<t_2 \ldots<t_n <\infty$, we have that:
$P((x_{t_1},x_{t_2},\ldots,x_{t_n})\in A)=P((y_{t_1},y_{t_2},\ldots,y_{t_n})\in A)$
(You can prove this IS a $\lambda-$ system by the fact that $P$ is a probability measure).
Then, the $\pi-$ system, which is the set of the "building-blocks" of $\mathcal B$ of the form
$A = (X_{t_1}\in A_1, X_{t_2}\in A_2, ..., X_{t_n}\in A_n)$, where each $A_i$ is a borel set in $R$
Use the fact that $Y$ is a modification of $X$ to show that the $\pi$ system is a subset of the $\lambda$ system, and you are done!
A: Consider sets $A_k:= \{\omega:X_{t_k}(\omega)\neq Y_{t_k}(\omega)\}$ for $k=1,\dots,n$. They all have $0$ probability. So 
$$
\mathbf P[\bigcup_{k=1}^n A_k]\leq \sum_{k=1}^n\mathbf P[A_k]=0
$$
which gives:
$$
\mathbf P[X_{t_k}=Y_{t_k}~for~all~k=1,\dots,n]=\mathbf P[\bigcap_{k=1}^nA_k^c] =1.
$$
So instead have the same distribution, you even have $(X_{t_1},X_{t_2},\dots,X_{t_n})$ as surely equal to $(Y_{t_1},Y_{t_2},\dots,Y_{t_n})$.
A: Since I could't understand the second answer in this post for a while (https://math.stackexchange.com/q/1709305 (version: 2016-03-23)) I would like to add my current understanding and derivation (as I cannot add comments, I use an answer window).
For simplicity of notation, let $\mathbf{X}\equiv(X_{t_1},\dots,X_{t_n}), \mathbf{Y}\equiv(Y_{t_1},\dots,Y_{t_n})$ and some $A\in\mathcal{B}(\mathbb{R}^{nd})$.
$$
|\mathbb{P}(\{\mathbf{X}\in A\})-\mathbb{P}(\{\mathbf{Y}\in A\})|=|\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})+\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})-\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})-\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})|=(1)
$$
by division of set on two disjont intersections. As on set $\{\mathbf{X=\mathbf{Y}}\}$ we must have $\{\mathbf{X}\in A\}\iff\{\mathbf{Y}\in A\}$, we can write
$$
\{\mathbf{X}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\}=\{\mathbf{Y}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\}\implies\\\
\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})=\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X=\mathbf{Y}}\})\\\ (1)=|\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})-\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})|\leq\mathbb{P}(\{\mathbf{X}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})+\mathbb{P}(\{\mathbf{Y}\in A\}\cap\{\mathbf{X\neq\mathbf{Y}}\})=(2)
$$
by triangle inequality. Both arguments of probabilities in last line are subsets of $\{\mathbf{X\neq\mathbf{Y}}\}$, so
$$
(2)\leq2\mathbb{P}(\{\mathbf{X}\neq\mathbf{Y}\})=0
$$
as a consequence, that processes are reciprocal modifications.
