(Proof Verification) The necessary and sufficient condition for $X:\mathbb{T}\times\Omega\longrightarrow\mathbb{R}$ to be a stochastic process. Recall that a stochastic process $\{X_{t}:t\in \mathbb{T}\}$ defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ leads to the mapping $\xi:\Omega\longrightarrow\mathbb{R}^{\mathbb{T}}$ which maps an outcome $\omega\in\Omega$ to the corresponding trajectory of the process, namely $\{t\mapsto X_{t}(\omega)\}\in\mathbb{R}^{\mathbb{T}}.$
Define the elementary cylinder as $$\{x\in\mathbb{R}^{\mathbb{T}}:x_{t_{1}}\in B_{1},\cdots, x_{t_{n}}\in B_{n}\},$$ for some $B_{1},\cdots, B_{n}\in\mathcal{B}(\mathbb{R})$.
Then, define the cylindrical $\sigma-$algebra by $\mathcal{B}(\mathbb{R}^{\mathbb{T}}):=\sigma(\text{elementary cylinders}).$
Now, I'd like to show that 

$X:\mathbb{T}\times\Omega\longrightarrow\mathbb{R}$ is a stochastic process if and only if $X$ seen as  $X:\Omega\longrightarrow\mathbb{R}^{\mathbb{T}}$ is $(\Omega,\mathcal{F})\longrightarrow(\mathbb{R}^{\mathbb{T}},\mathcal{B}(\mathbb{R}^{\mathbb{T}}))-$measurable. 

I showed the direction $(\Rightarrow)$ without problem, I "kind of" proved the converse direction but I had a question in the end:
Here is my proof for $(\Leftarrow)$:
Suppose $\xi:\Omega\longrightarrow\mathbb{R}^{T}$ is $(\Omega,\mathcal{F})\longrightarrow(\mathbb{R}^{\mathbb{T}},\mathcal{B}(\mathbb{R}^{\mathbb{T}}))-$measurable.  Then by hypothesis, for any elementary cylinder $E$, we have $\xi^{-1}(E)\in\mathcal{F}$. 
In particular, let $B\in\mathcal{B}(\mathbb{R})$, the elementary cylinder defined by $$E:=\{x\in\mathbb{R}^{\mathbb{T}}:x_{t_{1}}\in B, x_{t_{2}}\in\mathbb{R},\cdots,x_{t_{n}}\in\mathbb{R}\},$$ has the preimage 
\begin{align*}
\xi^{-1}(E)&=\{\omega\in\Omega:X_{t_{1}}(\omega)\in B, X_{t_{2}}(\omega)\in\mathbb{R},\cdots, X_{t_{3}}(\omega)\in\mathbb{R}\}\\
&=X_{t_{1}}^{-1}(B)\cap X_{t_{2}}^{-1}(\mathbb{R})\cap\cdots\cap X_{t_{n}}^{-1}(\mathbb{R})\\
&=X_{t_{1}}^{-1}(B)\cap\Omega\cap\cdots\cap\Omega\\
&=X_{t_{1}}^{-1}(B).
\end{align*}
But by hypothesis $\xi^{-1}(E)\in\mathcal{F}$, so we must have $X_{t_{1}}^{-1}(B)\in\mathcal{F}$. Since this holds for all $B\in\mathcal{B}(\mathbb{R})$, we are able to conclude that $X_{t_{1}}$ is measurable. 
We can do the similar things to each $X_{t_{i}}$ by just using $B\in\mathcal{B}(\mathbb{R})$ at $i^{th}$ coordinate and using $\mathbb{R}$ at all other coordinates. 
Thus, each $X_{t_{i}}:\Omega\longrightarrow\mathbb{R}$ is measurable, and thus $X$ is a stochastic process. 
The question I had is: my argument only showed $X_{t_{i}}$ is measuralble for finitely many of $t_{i}$, but a stochastic process $\mathbb{T}$ is not necessarily finitely indexed, i.e $\mathbb{T}$ may have infinite indices, for instance $\mathbb{T}=\mathbb{R}_{+}$. How could I adapt my proof to such a case? or I am having a misconception here so my current proof is okay? 
Thank you so much!
Edit 1:
Since it seems that I asked a trivial direction and said I could show the harder direction, I think it is better to post my proof of $(\Rightarrow)$.
Suppose $X:\mathbb{T}\times\Omega\longrightarrow\mathbb{R}$ is a stochastic process. Recall that a stochastic process $\{X_{t}:t\in \mathbb{T}\}$ defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ leads to the mapping $\xi:\Omega\longrightarrow\mathbb{R}^{\mathbb{T}}$ which maps an outcome $\omega\in\Omega$ to the corresponding trajectory of the process, namely $\{t\mapsto X_{t}(\omega)\}\in\mathbb{R}^{\mathbb{T}}.$
This mapping is $(\Omega,\mathcal{F})\longrightarrow(\mathbb{R}^{\mathbb{T}},\mathcal{B}(\mathbb{R}^{\mathbb{T}}))-$measurable, since the preimage of any elementary cylinder $$E=\{x\in\mathbb{R}^{\mathbb{T}}:x_{t_{1}}\in B_{1},\cdots, x_{t_{n}}\in B_{n}\},$$ is given by $$\xi^{-1}(E)=\{\omega\in\Omega:X_{t_{1}}(\omega)\in B_{1}, \cdots, X_{t_{n}}(\omega)\in B_{n}\}=X_{t_{1}}^{-1}(B_{1})\cap\cdots\cap X_{t_{n}}^{-1}(B_{n}).$$ But by hypothesis, each $X_{t_{i}}:\Omega\longrightarrow\mathbb{R}$ is measurable, and thus $X_{t_{i}}^{-1}(B_{i})\in\mathcal{F}$ for each $i$, and thus so is their intersection. 
Therefore, $\xi^{-1}(E)\in\mathcal{F}$. 
Thus $\xi$ is $(\Omega,\mathcal{F})\longrightarrow(\mathbb{R}^{\mathbb{T}},\mathcal{B}(\mathbb{R}^{\mathbb{T}}))-$measurable, as desired. 
 A: I will write $\mathrm{T}$ instead of $\mathbb{T}.$ Consider also a metrisable, separable space $\mathrm{E},$ with Borel $\sigma$-fiel $\mathscr{E}.$ Define $\mathrm{Z} = \mathrm{E}^\mathrm{T}$ and $\mathscr{Z} = \bigotimes\limits_{t \in \mathrm{T}} \mathscr{E},$ the product $\sigma$-field (separability of $\mathrm{E}$ allows deducing that $\mathscr{Z}$ is the Borel $\sigma$-field of $\mathrm{Z}$). Consider a probability space $(\Omega, \mathscr{F}, \mathbf{P}).$ We also write $p_t:\mathrm{Z} \to \mathrm{E}$ given by $p_t(z) = z(t) = z_t \in \mathrm{E}$ (the "projection onto $t$th factor" or the "evaluation at $t$th factor").
Theorem. A necessary and sufficient condition for a function $X:\Omega \to \mathrm{Z}$ to be measurable relative to $\mathscr{F}$ and $\mathscr{Z}$ is that for every $t \in \mathrm{T},$ the function $p_t \circ X = X_t:\Omega \to \mathrm{E}$ should be measurable relative to $\mathscr{F}$ and $\mathscr{E}.$
Proof. The necessity of the condition is obvious since each $p_t$ is $(\mathscr{Z}, \mathscr{E})$-measurable. To prove the sufficiency we have to show that for every $\mathrm{G} \in \mathscr{Z}$ the inverse image $X^{-1}(\mathrm{G})$ belongs to $\mathscr{F}.$ Denote then by $X^{-1}(\mathscr{Z})$ the set of all $X^{-1}(\mathrm{G})$ as $\mathrm{G}$ runs on $\mathscr{Z}.$ Thus, we aim at showing $X^{-1}(\mathscr{Z}) \subset \mathscr{F}.$ Denote by $\mathscr{Y}$ the set of all $\mathrm{G} \in \mathscr{Z}$ such that $X^{-1}(\mathrm{G}) \in \mathscr{F}.$ The fact that $\mathscr{Z}$ is the product $\sigma$-field means that $\mathscr{Z} = \sigma(p_t^{-1}(\mathrm{L}); \mathrm{L} \in \mathscr{E}, t \in \mathrm{T}).$ The hypothesis give $X^{-1}(p_t^{-1}(\mathrm{L})) = X_t^{-1}(\mathrm{L}) \in \mathscr{F},$ thus $\mathscr{Y}$ contains $p_t^{-1}(\mathrm{L})$ for $t \in \mathrm{T}$ and $\mathrm{L} \in \mathscr{E}.$ It also easy to see that $\mathscr{Y}$ is a $\sigma$-algebra. Thus $\mathscr{Y} = \mathscr{Z}.$ Q.E.D.
