Exercise 6.9 in Rudin's RCA (Real and Complex Analysis)
This counter example that shows that the answer is no in general:
- For each integer $n$ define $g_n$ to be a nonnegative continuous function on $[0,1]$ which equals $0$ except on intervals of the form $I_{n,k}=\big[\frac{k}{n},\frac{k}{n}+\frac{1}{2^n}\big]$, $0\leq k<n$, where $$\int_{I_{nk}}g_n=\frac1n$$ One can for instance use piecewise linear functions that take large values at points $k/n$, $0\leq k<n$.
- This sequence converges to $0$ in measure: $m\big(|g_n|>\varepsilon\big)\leq m\big(g_n\neq0\big)\leq n 2^{-n}\xrightarrow{n\rightarrow\infty}0$ for all $\varepsilon>0$.
- One can then take a subsequent $g_{n'}$ which converges to $0$ $m$-a.s. For sake of simplicity, let us denote the subsequence as $g_n$ (abuse of notation if you will)
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Claim: Let $f\in\mathcal{C}[0,1]$.
$$\lim_n\int f g_n\,dm = \int f\,dm\tag{1}\label{one}$$ One can appeal to uniform continuity to get for any $\varepsilon >0$, a $\delta>0$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. Taking $n$ sufficiently large (e.g. $n>\frac{1}{\delta}$)
$$ \Big(f\big(\frac{k}{n}\big) -\varepsilon\Big)g_n\mathbb{1}_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]}\leq f g_n\mathbb{1}_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]}\leq \Big(\varepsilon + f\big(\frac{k}{n}\big)\Big)g_n\mathbb{1}_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]} $$ Integration gives $$ \Big( f\big(\frac{k}{n}\big) -\varepsilon\Big)\frac{1}{n}\leq \int_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]} f g_n\,dm \leq \Big(\varepsilon + f\big(\frac{k}{n}\big)\Big)\frac{1}{n} $$ Adding over all $0\leq k<n$ results in $$ \Big|\int_I fg_n\,dm -\sum^{n-1}_{k=0}f\big(\frac{k}{n}\big)\frac{1}{n}\Big|\leq\varepsilon $$ As $f$ is continuous, $\sum^{n-1}_{k=0}f\big(\frac{k}{n}\big)\frac{1}{n}\xrightarrow{n\rightarrow\infty}\int_If\,dm$ and the claim follows.
- This gives a negative answer to the question in Rudin's problem by taking $\mu=m$.