Infinite Product Measures and Kolmogorov Extension — Examples

The countable product theorem supplies the measure with the displayed finite product marginals.

Each specified coordinate atom has mass $1\mathrm{/}2$, so the finite rectangle has mass ${\prod }_{j=1}^k(1\mathrm{/}2)=2^{-k}$; the coordinates are independent by the coordinate-independence corollary.

Apply [F1] with $(E_n,{\mathcal{E}}_n,{\mu }_n)=(S,\mathrm{\Sigma },\nu )$ for every $n\in \mathbb{N}$. It gives the canonical probability on the cylinder sigma-algebra of $S^{\mathbb{N}}$. By [F2], its coordinate maps $X_n(x)=x_n$ are independent and each has law $\nu$.

Its finite-family conclusion is independence, while its one-coordinate conclusion is the common marginal; together these are the definition of i.i.d.

The countable-product theorem supplies the product probability under the stated choice principles. The coordinate-independence corollary applies to this sequence of Bernoulli laws and gives independence.

$\mathbb{P}(X_0=1)=1\mathrm{/}2$ while $\mathbb{P}(X_1=1)=1\mathrm{/}3$. Hence their laws differ, so the independent family is not identically distributed.

The countable product theorem gives the probability measure with fair Bernoulli coordinate marginals. The coordinate-independence corollary makes the $X_n$ independent. Thus the signs ${\xi }_n$ are independent, with $\mathbb{P}({\xi }_n=1)=\mathbb{P}({\xi }_n=-1)=1\mathrm{/}2$.

For any finite list of distinct increment times, their joint law is the uniform law on the corresponding sign vectors. The process coordinates are the partial sums of these increments, not the increments themselves.

The finite uniform laws are compatible under deletion of coordinates, so the standard-Borel extension theorem applies.

The canonical-process corollary identifies each prescribed finite joint law. Thus finite coordinate events are measurable and have their product probabilities; no assertion is made about arbitrary path events.

For fixed $t$, $\{X_t\mathrm{\ne }{Y}_t\}=\{U=t\}$ is measurable and has Lebesgue probability $0$, so $Y$ is a modification of $X$ under the stated diagonal convention.

For every outcome $u$, choosing $t=u$ gives $Y_t(u)=1\mathrm{\ne }0=X_t(u)$. Thus the equality-for-all-times event is empty, not probability one, and the processes are not indistinguishable.

Each displayed coordinate has Bernoulli$(1\mathrm{/}2)$ law in both pairs.

But $\mathbb{P}(Z=Z)=1$ for the first pair, whereas $\mathbb{P}(Z=W)=1\mathrm{/}2$ for the independent pair. Their two-dimensional laws differ, refuting the statement.

If $A$ were cylinder-measurable, the countable-coordinate lemma would give a countable $J\subseteq I$ determining it.

Choose $i\in I\setminus J$. The all-zero path and the path with ones on $I\setminus J$ agree on $J$, but only the latter belongs to $A$; this contradicts determination by $J$.

Hence $A$ is not cylinder-measurable and $f$ is not a real random variable on the canonical cylinder probability space.