Let $f$ be a complex-valued continuous function on $\mathbb{R}_+$ with compact support and let $g, h$ be two complex-valued continuous functions on $\mathbb{R}_+$ such that $g$ is bounded and $|h(t)|$=1. Let $0 \leq s < t < \infty$.
Find a limit as $n \to \infty$ of
$\prod \limits_{j=k+1}^{r} \left[ \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}+f(\frac{j}{n})h(\frac{j}{n})\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}\right] \times \prod \limits_{j\notin \{k+1, \ldots, r \}}\left(1+\frac{1}{n}f(\frac{j}{n})\right) $
where $\frac{k}{n}< s \leq \frac{k+1}{n} \leq \frac{r}{n} \leq t < \frac{r+1}{n}$.
It should be of the form $\exp\left[\int_{0}^{\infty}f(x)dx + \int_{s}^{t} \left(-\frac{1}{2}|g(x)|^2 + f(x)(h(x)-1)\right) dx\right]$.
Hints are mean value theorems and Plancherel theorem, but I have no idea how to apply it.
Thank you for any hints and answers.
There is a $\frac1n$ missing before the term $f(\frac{j}{n})h(\frac{j}{n})$.
Taking logarithms turns this product into a Riemann sum plus some terms which are negligible in the limit. For example, observe that, since $f$ has compact support, we can assume that $f(x)$ vanishes for $x> L$, for some fixed $L>t$. Then the product over $j\notin\{k+1,\ldots,r\}$ can be cut off at $Ln$, so \begin{eqnarray*} &&\log \prod_{j\notin\{k+1,\ldots,r\}} (1+\frac1n f(\frac jn))\\ &=&\sum_{j\notin\{k+1,\ldots,r\},\ j\le Ln} \log (1+\frac1n f(\frac jn))\\ &=&\sum_{j\notin\{k+1,\ldots,r\},\ j\le Ln} (\frac1n f(\frac jn) + O(\frac{1}{n^2})),\\ \end{eqnarray*} where, since $|f|$ is bounded on ${\Bbb R}_+$, the $O()$ term is uniform in $j$ $$ =O(\frac1n) + \sum_{j\notin\{k+1,\ldots,r\},\ j\le Ln} \frac1n f(\frac jn). $$ The first term vanishes in the limit and the second is a Riemann sum whose limit is $$\int_0^s f(x) \, dx+\int_t^L f(x) \, dx.$$
The term in the other part of the product, $$ \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}+\frac1nf(\frac{j}{n})h(\frac{j}{n})\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2},$$ factors as $$ \sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}\left(1+\frac1nf(\frac{j}{n})h(\frac{j}{n})\right).$$ You can then deal with in it the same way, giving a logarithmic limit of $$ \int_s^t -\frac12 |g(x)|^2 + f(x)h(x) \, dx. $$ Adding and exponentiating then gives the desired result.
Re the comments below, suppose that you replaced the term in the first product by $$ T_{jn}:=\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}+\frac1n \left(h(\frac{j}{n})+f(\frac{j}{n})h(\frac{j}{n})\sqrt{1- \frac{1}{n}| g(\frac{j}{n})|^2}\right).\ \ (1)$$ Then, use the Taylor expansion $$ \sqrt{1-\frac{1}{n} |g(\frac{j}{n})|^2}=1-\frac{1}{2n} |g(\frac{j}{n})|^2+O(\frac{1}{n^2}).\qquad (2) $$ Since $g$ is bounded, the $O()$ term is uniform in $j$. Therefore, substituting $(2)$ into $(1)$, $$ T_{jn}=1-\frac{1}{2n} |g(\frac{j}{n})|^2+\frac1n h(\frac jn) + \frac1n f(\frac jn) h(\frac jn) + O(\frac{1}{n^2}), $$ and so $$ \log T_{jn}=-\frac{1}{2n} |g(\frac{j}{n})|^2+\frac1n h(\frac jn) + \frac1n f(\frac jn) h(\frac jn) +O(\frac{1}{n^2}).\qquad (3) $$ Since $f$, $g$ and $h$ are all bounded on the interval being considered, the $O()$ term is still uniform in $j$ in both of these expressions. Then \begin{eqnarray*} &&\log \prod_{k+1\le j\le r} T_{jn}\\ &=&\sum_{k+1\le j\le r} \log T_{jn}\\ &=&O(\frac 1n)+\sum_{k+1\le j\le r} -\frac{1}{2n} |g(\frac{j}{n})|^2+\frac1n h(\frac jn) + \frac1n f(\frac jn) h(\frac jn),\\ \end{eqnarray*} using $(3)$, and as before the sum on the right-hand side is a Riemann sum.