Vtyjkyuk

This question is a slight variation of Rudin's Real and Complex Analysis (Third Edition 1987) Chapter 6 Q9. It has been asked here Mutually Singular measures.

I was wondering if instead of restricting the function f to just C$[0,1]$, we allow $f$ to be in $L^infty $ i.e the dual space of $L^1$. Precisely:

Suppose that $g_n$ is a sequence of positive continuous functions on I=$[0,1]$, $mu$ is a positive Borel measure on I, $m$ is the standard Lebesgue measure, and that

(i) $lim_n→inftyg_n(x)=0$ a.e. $[m]$

(ii) $int _I g_ndm=1$ for all n,

(iii) $lim_n→inftyint _I fg_ndm=int _I f dmu$ for every $mathbff∈L^infty(I)$.

Does it follow that the measures μ and m are mutually singular?

$textbfThoughts:$

I think it should be true.

Using Radom Nikodym Theorem we can write $dmu=hdm+dkappa$, where $kappa$ is mutually singular to $m$

Now let $B$ be the set on which $m$ is supported thus $kappa(B)=0$ and $m(Ibackslash B)=0$.

On $B$, $g_n$ converges weakly to $h$ and since $g_n$ converges pointwise to $0$, using the answer here

Real Analysis Convergence question

We get that $h=0$ a.e $B$ and thus also a.e I. Hence $mu$ and m are mutually singular.

Am I correct in doing so? If so, it is expected but at the same time interesting.

Thanks in advance.