Does $\prod_{i=1}^N (1-f(x_i)\Delta x_i+o(\Delta x_i))\to e^{-\int_0^1f(x)dx}$ as $n\to\infty$?

Question

This is a problem that came up in my research and I was wondering if someone good at analysis could help me.

If $0=x_0<\dots<x_N=1$, $f:[0,1]\to\mathbb{R}$ is integrable,then does $P_N=\prod_{i=1}^N (1-f(x_i)\Delta x_i+\varepsilon_i)$ where $\varepsilon_i$ is $o(\Delta x_i)$ as $\Delta x_i\to0$.

I want to say that $P_N\to e^{-\int_0^1f(x)dx}$ as $N\to\infty$, but I am unsure if it is true. If it is not true in general, then under what conditions on $e$ does it hold? Does it hold if $\varepsilon_i$ is $o(\Delta x_i)$ as $\Delta x_i\to0$ uniformly in $i$?

I tried to take logs, which leads to,

$$\log P_N=\sum_{i=1}^N\log (1-f(x_i)\Delta x_i+e_i)\approx-\sum_{i=1}^N f(x_i)\Delta x_i +\sum_{i=1}^N e_i.$$

I am not sure how to deal with the error term.

Answer 1

Consider the logarithm of the expression, $$ \ln\prod^N_{i=1}(1−f(x_i)Δx_i)=\sum^N_{i=1}\ln(1−f(x_i)Δx_i) =-\sum^N_{i=1}f(x_i)Δx_i -\frac12(f(x_i)Δx_i)^2+... $$ so that the first term is a Riemann/Darboux sum and the following expressions go to zero like $(\max|Δx_i|)^{k-1}$.