Weak Limit of Measures Mutually Singular wrt Lebesgue Measure I'm stuck on the following qual problem:
Let $\{h_{n}\}$ be a sequence of positive continuous functions on the unit cube $Q$ in $\mathbb{R}^{d}$ satisfying the following conditions:


*

*$\lim_{n\rightarrow\infty}h_{n}(x)=0$ $m$-a.e. ($m$ denotes the Lebesgue measure on $Q$)

*$\int_{Q}h_{n}dx=1$ $\forall n$

*$\lim_{n\rightarrow\infty}\int_{Q}fh_{n}dx=\int_{Q}fd\mu$ for every continuous function $f$ on $Q$.


Prove that $\mu\perp m$ or give a counterexample.
My intuition suggests to me that $\mu\perp m$ since $h_{n}\rightarrow 0$ a.e. and therefore must become very large on sets of small Lebesgue measure; however, I'm struggling to prove my guess. My thought was to write the $\int_{Q}fd\mu=\int_{Q}fh dx+\int_{Q}fd\nu$, where $hdx+\nu=\mu$ is the Lebesgue decomposition of $\mu$ and then show that $h=0$ $m$-a.e, or equivalently $\int_{Q}fhdx=0$ for every continuous $f$. But I have been unable to do this. Any suggestions?
 A: The flaw in reasoning that $\mu$ must be singular because $h_n$ is very small except on a very small set is that very small set can be somewhat uniformly distributed (at the right resolution).
Counterexample in one dimension:
Define $$I_{n,j}=[j/n,(j+2^{-n})/n].$$  Let $\phi_{n,j}$ be a continuous function supported on $I_{n,j}$ with $\phi_{n,j}\ge0$ and $$\int\phi_{n,j}=1/n.$$ Set $$E_n=\bigcup_{j=0}^{n-1}I_{n,j}.$$Let $$h_n=\sum_{j=0}^{n-1}\phi_{n,j}.$$Then $h_n\ge0$, $\int h_n=1$, and $$\int fh_n\to\int_0^1f(x)\,dx$$for every $f\in C([0,1])$. (Hint: $f$ is uniformly continuous. The idea behind the example is that $\int fh_n$ is morally equivalent to a Riemann sum for $\int f$.)
Since $h_n=0$ on $[0,1]\setminus E_n$ and $\sum m(E_n)<\infty$ it follows that $h_n\to0$ almost everywhere.
If $h_n\ge0$ is not positive enough, let $g_n=(1-1/n)h_n+1/n$.
Exercise: Suppose $\mu$ is a Borel probability measure on $[0,1]$. Show that there exist $h_n\ge0$ such that $h_n\to0$ almost everywhere but $\int fh_n\to\int f\,d\mu$ for every $f\in C([0,1])$.
