Infinite product of probability measures is a premeasure This is an exercise from Real Analysis by Stein and Shakarchi (Chapter 6, Exercise 15).

Given infinitely many measure spaces $(X_i, \mathcal M_i, m_i)$, each of which has measure 1, one can define an algebra on the product space consisting of all finite unions of the “cylinders”, by which we mean rectangles of the form $E_1 \times E_2 \times \cdots$, where $E_i$ belong to $\mathcal M_i$ and all but finitely many of $E_i$ are equal to $X_i$. Then define $m(E_1 \times E_2 \times \cdots) = m_1(E_1)\,m_2(E_2)\cdots$. How does one prove that $m$ is a premeasure on the algebra defined above?

One only needs to check the equality in the definition of premeasure, but it seems a subtle problem of the exchange of summation and limit progress is involved, which can be easily ignored without carefulness.

I would like some hints or any reference book about it.
This is proven as Theorem 3.5.1 in Measure Theory, Volume 1 by Vladimir Bogachev (WorldCat link if you want to find a library copy).

An overarching theme of the proof is that each cylinder set is essentially an element of finite product $\sigma$-algebras; the countably infinitely many factors $X_i$ tacked on at the end act essentially as filler.

Hint.

First prove the following lemma: Let $(X,\mathcal M, \mu)$ and $(Y,\mathcal N, \nu)$ be measure spaces. Suppose $A \in \mathcal M \otimes \mathcal N$ satisfies $(\mu \times \nu)(A) &gt; t &gt; 0$. For $x \in X$, we have the sections/slices $A^x := \{y \in Y : (x,y) \in A\}$. Let $B := \{x \in X : \nu(A^x) &gt; t/2\}$. Then $B \in \mathcal M$ and $\mu(B) &gt; t/2$.

To establish the main result, it suffices to show that if $(A_k)_{k=1}^\infty$ is a sequence of cylinder sets for which there exists a constant $\epsilon &gt; 0$ such that $m(A_k) &gt; \epsilon$ for all $k$, then $\bigcap_{k=1}^\infty A_k \subset \prod_{j=1}^\infty X_j$ has at least one element. The lemma is helpful here.



There's another proof outlined in Sam Drury's class notes for his Math 355 course at McGill under Section 4.6, "Infinite products of probability spaces". You can find my commentary on the proof in an old revision of this answer.
The trick is that we know the union must have only finitely many non-trivial co-ordinates and that the unions are increasing.

To prove $m$ is a pre-measure, you only need to prove the summation condition for disjoint events $\{A_i\}_{i \ge 1}$ where $\cup_i A_i$ lies in the algebra. Only way $\cup_i A_i$ lies in the algebra is if $$\cup_i A_i = E_{i_1} \times E_{i_2 } \times \ldots \times E_{i_k}$$

Let $E_{i_j}^{(N)}$ be the $i_j$th coordinate for $\cup_{i=1}^N A_i$.

which means for all $\epsilon &gt;0$, for all large enough $N$ , $$m (E_{i_j} \setminus E_{i_j}^{(N)}) &lt; \epsilon$$
for all $j=1,\ldots, k$.Hence 
$$ m(\cup_iA_i) -  m(\cup_{i=1}^N A_i) = m (\cup_iA_i \setminus \cup_{i=1}^N A_i)  &lt; \prod_{j=1}^k m(E_{i_j} \setminus E_{i_j}^{(N)}) &lt; \epsilon^k$$

since in $\cup_iA_i \setminus \cup_{i=1}^N A_i$, there are finitely many non-trivial co-ordinates and we can replace the measures of all co-ordinates except $i_1,\ldots, i_k$ by $1$ for the upperbound.