This is essentially an extensive comment on the post by Pete L. Clark.
We can define a product of a function with real values $(a_i)_{i\in I}$ in the following way (an elaboration on the definition by Pete): Let $\mathcal{F}$ be the set of finite subsets of $I$. It is a directed set under set inclusion. We define a net $\pi:\mathcal{F}\to\mathbb{R}$ by $\pi(F)=\times_{i\in F}a_i$. We let $\times_i a_i$ be the limit of the net $\pi$ if it exists. It is worth pointing out that this notion of convergence differs from the usual infinite product of a sequence of numbers in which convergence to zero is ruled out, so that convergence becomes equivalent to convergence of the series of logarithms of the factors. Essentially, the limit is then defined to be undefined when it would be zero otherwise. This makes little sense in probability theory, where we often want to show that a sequence of independent events has probability $0$. Also, the product defined here is "absolute", it does not depend on any order on the index set $I$.
Fact: If $(a_i)\in [0,1]^I$, then $\times_i a_i$ exists. If, moreover, $\times_i a_i>0$, then there exists a countable subset $C\subseteq I$ with $a_i=1$ for all $i\in I\backslash C$.
Proof: From the obvius fact that if $P\geq 0$ and $0\leq a_i\leq 1$ one has $Pa_i\leq P$, it follows that the net $\pi$ is nonincreasing. It is also obviously bounded below by $0$, so $\times_i a_i$ exists as the limit of a nonincreasing net that is bounded below. Assume now that $\times_i a_i>0$. For each positive natural number $n$, let $F_n=\{ i\in I:a_i<1-1/n \}$. For every $n$, $F_n$ is finite, so $\{i\in I:a_i<1\}=\bigcup_n F_n$ is countable.
Fact: If $I$ is countable and for every $i\in I$, $Y_i$ is a measurable subset of $X_i$, then $\prod_i Y_i$ is in the product $\sigma$-algebra.
Proof: W.l.o.g., let $I=\mathbb{N}$. For every $n$, the set $S_n=X_1\times\ldots X_{n-1}\times Y_n\times X_{n+1}\times\ldots=\pi_n^{-1}(Y_n)$ is by definition in the product $\sigma$-algebra. Now $\prod_{n\in\mathbb{N}}Y_n=\bigcap_{n=1}^\infty S_n$ which is in the $\sigma$-algebra as the countable intersection of measurable sets.
By a similar argument, one can show that for a general index set $I$, every measurable set can be written (subject to the usual labeling issues) as $\prod_{i\in C}Y_i\times\prod_{i\in I\backslash{C}}X_i$ with $C$ being a countable subset of $I$ and $Y_i$ an arbitrary measurable subset of $X_i$. So every measurable set in the product $\sigma$-algebra is determined by countably many coordinates. A useful Lemma for thinking about this and related issues is the following:
Lemma: Let $X$ be a set and $\mathcal{Y}$ be a family of subsets of $X$ and let $B\in\sigma(\mathcal{Y})$. Then there exists a countable family $\mathcal{C}\subseteq\mathcal{Y}$ such that $B\in\sigma(\mathcal{C})$.
Proof: Just verify that the family of all sets constructed by countable subfamilies forms a $\sigma$-algebra.
As a last remark: Tychonoff's theorem has a direct application in studying probabilities on infinite product spaces. The crucial step in proving the Daniell-Kolmogorov Extension Theorem is showing that the resulting probability measure is not just finitely additive but countably additive on the product algebra. For this one uses regularity with respect to the compact sets in the product topology (see here for a general approach). This is also the reason why the theorem does not hold for arbitrary product spaces (so the independent product of probability spaces is quite special).