Infinite product - pushing the limit (w.r.t. n) inside when the upper index depends on n as well Let $f_1,f_2,...$ be a non-negative functions s.t. $f_n$ converge uniformly to $f$ on compact intervals and let $a_1,a_2,...$ be real numbers s.t. $\forall j\in\mathbb{N}, 0<a_j<1$.
In addition assume that
$$
\prod_{j=1}^\infty f(a_j)\in\mathbb{R} 
$$
Under which conditions we have the following?
$$
\lim_{n\to\infty}\prod_{j=1}^n f_n(a_j)=\prod_{j=1}^\infty f(a_j)
$$
 A: Provisional (fairly basic) response to the updated question:
It may depend on the rate of convergence.
Using Lebesgue’s theory, if we have that there exists a summable $L^1(\Bbb N)$ sequence $b_j$ such that $|\ln f_n(a_j)|\le b_j$ for all $n,j$, then the dominated convergence theorem gives you the result. Alternatively, if $\ln f_n(a_j)\to\ln f(a_j)$ monotonically, then the monotone convergence theorem - also from measure theory - saves the day.
My attempt at using real analysis (pre-measure theory) was not very successful. While uniform convergence is successful with, say, the Riemann integral over compact intervals, it doesn’t immediately help us when dealing with infinite series, as far as I know, because there is the trouble with the infinity of the measure space, analogous to a domain of integration, unlike when integrating over bounded intervals.
We know that $f_n\rightrightarrows f$ on $[0,1]$, and that $\sum_{j=1}^\infty f(a_j)$ is convergent. However, $\ln f_n$ might not uniformly converge to $\ln f$ since $\ln$ is not uniformly continuous over $(0,\infty)$. To ensure $\ln f_n\rightrightarrows\ln f$, I think we would require some very explicit and rapid bounds, or require that, for some $d>c>0$, $c<f_n(a_j)<d$ (because then $\ln$ is uniformly continuous over $[c,d]$ and, as far as the $a_j$ are concerned, $\ln f_n$ uniformly tends to $\ln f$).
If we do have $\ln f_n(a_j)\rightrightarrows\ln f(a_j)$ then I believe we can argue as follows (with further assumptions):

Let $L:=\sum_{j=1}^\infty\ln f(a_j)$ (assume finite, i.e. the product cannot be zero) and $L’:=\lim_{n\to\infty}\sum_{j=1}^\infty\ln f_n(a_j)$ (assume existence as a finite limit). For all $0<\varepsilon$, there exists $N\in\Bbb N$ so that the $n$th partial term of each limit is $\varepsilon$-close to $L,L’$ for $n>N$. Let $N_{\varepsilon}>N$ be the smallest such natural where $\|f_n-f\|\lt\varepsilon$ for $n>N_{\varepsilon}$. For such large $n$, we can estimate: $$\begin{align}|L-L’|&\le\left|L-\sum_{j=1}^n\ln f(a_j)\right|+\left|\sum_{j=1}^n[\ln f(a_j)-\ln f_n(a_j)]\right|+\left|L’-\sum_{j=1}^n\ln f_n(a_j)\right|\\&\lt(2+n)\varepsilon\end{align}$$However, we need to assume that $N_{\varepsilon}\varepsilon\to0$ as $\varepsilon\to0$, in order for this argument to succeed.

If I remember correctly, we can extract a rapid Cauchy subsequence to get, along this subsequence, $N_{\varepsilon}\varepsilon\to0$.

EDIT: The below answer was in response to a previous edition of the question.

A counterexample.
I'll index from $j=1$ for convenience. Let $a_{j,n}=j^{-1/n}\in(0,1]$ for all $(j,n)\in\Bbb N^2$. Then: $$\pi_n:=\prod_{j=1}^na_{j,n}=\left(\frac{1}{1}\frac{1}{2}\frac{1}{3}\cdots\frac{1}{n}\right)^{1/n}=(n!)^{-1/n}$$And $\lim_{n\to\infty}\pi_n=0$ can be checked, for example you can read this post. However, $\lim_{n\to\infty}a_{j,n}\equiv1$ for all $j$, so we get: $$0=\lim_{n\to\infty}\prod_{j=1}^na_{j,n}\neq\prod_{j=1}^\infty(\lim_{n\to\infty}a_{j,n})=1$$
This is a good example of why we must be very careful, in principle, before exchanging limits.
By taking logarithms on both sides, and appealing to the continuity of the exponential, to answer your question in general we are seeking to understand when: $$\lim_{n\to\infty}\sum_{j=1}^n\ln a_{j,n}=\sum_{j=1}^\infty(\lim_{n\to\infty}\ln a_{j,n})$$Allowing either side to exist as a divergent infinite limit (in my counterexample, the limit of the left hand side is $-\infty$). If you have an opportunity to employ the dominated convergence theorem, or some similar result, then you can exchange the limits. Otherwise, maybe you can't. The question of: "when can the limit of the series be the series of the limit?" is very general, but there is a lot of theory and work already done on the subject that you can investigate.
