If $f,g$ are continuous and $g$ is $1$-periodic, $\int_0^1 f(x)g(nx)dx \xrightarrow[n\to\infty]{} \int_0^1 f \int_0^1 g$

We have

$$\begin{align} \int_0^1f(x)g(nx)dx &= \frac 1n\int_0^n f\left(\frac{x}{n}\right)g(x)dx = \frac{1}{n}\sum_{k=0}^{n-1}\int_k^{k+1}f\left(\frac{x}{n}\right)g(x)dx \\ &= \frac 1n\sum_{k=0}^{n-1}\int_0^1 f\left(\frac{x+k}{n}\right)g(x)dx. \end{align}$$

Now use that for $n$ large enough by uniform continuity of $f$ and boundedness of $g$ we can get $f(\frac{x+k}{n})g(x)$ in an $\epsilon$-range of $f(\frac kn)g(x)$. Summed up and divided by $n$ we remain in an $\epsilon$-range if we replace $f(\frac{x+k}{n})$ in the integral by $f(\frac kn)$.

But then we have

$$ \frac 1n\sum_{k=0}^{n-1}\int_0^1f\left(\frac{k}{n}\right)g(x)dx = \frac 1n\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right)\int_0^1g(x)dx. $$

Take it from there.

To complement LeBtz's excellent answer above, let's try a more heuristic way of figuring out what to do. This may not be the shortest, but should (i) give an intuition of what is going on; and (ii) show that is it not "magic" with the proof coming out of the blue, but that there is some process to arrive at the result.

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Since the assumption on $g$ is $1$-periodicity, we want to make appear things like $g(u+k)$ for $k\in\mathbb{N}$, so that we can simplify. So let us break $[0,1]$ in many segments of length $\frac{1}{n}$, so that $g(nx)$ becomes something we want: $$\begin{align} \int_0^1 f(x)g(nx)dx &= \sum_{k=0}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} f(x)g(nx)dx = \sum_{k=0}^{n-1} \int_{0}^{\frac{1}{n}} f\left(u+\frac{k}{n}\right)g\left(n\left(u+\frac{k}{n}\right)\right)du \\ &=\sum_{k=0}^{n-1} \int_{0}^{\frac{1}{n}} f\left(u+\frac{k}{n}\right)g\left(nu+k\right)du =\sum_{k=0}^{n-1} \int_{0}^{\frac{1}{n}} f\left(u+\frac{k}{n}\right)g(nu)du \end{align} $$ That's good, we got somewhere. Not quite there, but still... we can get a little more done now by swtiching integral and sum, and observing that after simplification the term $g(nu)$ does not depend on the summation index $k$ anymore: $$\begin{align} \int_0^1 f(x)g(nx)dx &= \sum_{k=0}^{n-1} \int_{0}^{\frac{1}{n}} f\left(u+\frac{k}{n}\right)g(nu)du = \int_{0}^{\frac{1}{n}} du \sum_{k=0}^{n-1} f\left(u+\frac{k}{n}\right)g(nu)\\ &= \int_{0}^{\frac{1}{n}} du\, g(nu) \sum_{k=0}^{n-1} f\left(u+\frac{k}{n}\right) \end{align} $$ Not too bad. Let's continue — eventually, we want integrals from $0$ to $1$, so let us switch back via a change of variables: $$\begin{align} \int_0^1 f(x)g(nx)dx &= \int_{0}^{\frac{1}{n}} du\, g(nu) \sum_{k=0}^{n-1} f\left(u+\frac{k}{n}\right) = \int_{0}^{1} dx\, g(x) \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{x}{n}+\frac{k}{n}\right) \end{align} $$

That looks promising. Why? Well, if the last term had been $\frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right)$ instead, we would have had a Riemann sum for $f$, whose limit as $n\to \infty$ is exactly $\int_0^1 f$... this is nice. But there is that extra $\frac{x}{n}$ which is preventing that from directly applying: this is where the (uniform, since on a bounded closed interval) continuity of $f$ should come into play, as $\frac{x}{n}\xrightarrow[n\to\infty]{}0$.

• By the above, $$\begin{align} \int_{0}^{1} dx\, g(x) \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) = \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right)\int_{0}^{1} g(x)dx \xrightarrow[n\to\infty]{} \int_{0}^{1} f(x)dx \int_{0}^{1} g(x)dx \end{align} \tag{1} $$

• Let us control the difference, call it $\Delta_n$: $$\begin{align} \Delta_n &= \left\lvert \int_{0}^{1} dx\, g(x) \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) - \int_{0}^{1} dx\, g(x) \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{x}{n}+\frac{k}{n}\right) \right\rvert \\&= \left\lvert \int_{0}^{1} dx\, g(x) \frac{1}{n}\sum_{k=0}^{n-1} \left( f\left(\frac{k}{n}\right) - f\left(\frac{x}{n}+\frac{k}{n}\right) \right) \right\rvert \\ &\leq \int_{0}^{1} dx\, \lvert g(x)\rvert \frac{1}{n}\sum_{k=0}^{n-1} \left\lvert f\left(\frac{k}{n}\right) - f\left(\frac{x}{n}+\frac{k}{n}\right) \right\rvert \end{align} $$

  For convenience, write $\lVert g\rVert_\infty\stackrel{\rm def}{=} \max_{x\in[0,1]} \lvert g(x)\rvert$ (this exists, as $g$ is continuous). Fix any $\varepsilon > 0$, and — $f$ is uniformly continuous too — let $N_\varepsilon$ such that, for all $n\geq N_\varepsilon$, $\lvert f(x)-f(y) \rvert \leq \frac{\varepsilon}{\lVert g\rVert_\infty+1}$ whenever $x,y\in[0,1]$ satisfy $\lvert x-y\rvert \leq \frac{1}{n}$. (The $+1$ is just to avoid dividing by zero if $g$ is identically zero.)

  Then we are good: for any $n\geq N_\varepsilon$, since $\frac{x}{n} \leq \frac{1}{n}$, we get

  $$\begin{align} \Delta_n &\leq \int_{0}^{1} dx\, \lvert g(x)\rvert \frac{1}{n}\sum_{k=0}^{n-1} \left\lvert f\left(\frac{k}{n}\right) - f\left(\frac{x}{n}+\frac{k}{n}\right) \right\rvert\\ &\leq \int_{0}^{1} dx\, \lVert g\rVert_\infty \frac{1}{n}\sum_{k=0}^{n-1} \frac{\varepsilon}{\lVert g\rVert_\infty+1} = \varepsilon \frac{\lVert g\rVert_\infty}{\lVert g\rVert_\infty+1}\\ &< \varepsilon \end{align}$$ and since $\varepsilon>0$ was arbitrary, we just showed that $$\Delta_n \xrightarrow[n\to\infty]{} 0. \tag{2} $$

Combining (1) and (2) gives the limit:

$$ \int_0^1 f(x)g(nx)dx = \int_{0}^{1} dx\, g(x) \frac{1}{n}\sum_{k=0}^{n-1} f\left(\frac{k}{n}\right) +\Delta_n \xrightarrow[n\to\infty]{} \int_{0}^{1} f(x)dx \int_{0}^{1} g(x)dx+0 $$