TITLE: Fourier coefficient of complex measure (Exercise in Rudin)
QUESTION [5 upvotes]: Problem: Suppose $\mu$ is a complex Borel measure on $[0,2\pi)$, and define the Fourier coefficients of $\mu$ by
  $$\hat{\mu}(n)=\int_{[0,2\pi)}e^{-int}d\mu(t),\quad n\in\mathbb{Z}.$$
  Assume that $\hat{\mu}(n)\to 0$ as $n\to+\infty$ and prove that then $\hat{\mu}(n)\to 0$ as $n\to-\infty$.

Source: Exercise 7 in Chapter 6 of Rudin's Real and Complex Analysis.
Attempt: Rudin gives the hint that we can reduce to the case $\mu\geq0$. I was able to do this without too much difficulty. The idea is that we can replace $d\mu$ by $pd\mu$ if $p$ is any trigonometric polynomial, and hence by $fd\mu$ if $f$ is continuous, and therefore if $f$ is a bounded Borel function (by Lusin's theorem). Hence, since $d\mu=hd|\mu|$ for some $h\in L^1(|\mu|)$ with $|h|=1$, and since bounded functions are dense in $L^1(|\mu|)$, we can replace $d\mu$ by $d|\mu|$.
But then I am not sure how to proceed. We know that a bounded Borel measure on an interval $I$ is regular (special case of Theorem 2.18). And $|\mu|$ is bounded since $\mu$ is a complex measure. So $|\mu|$ is regular. Hence:

Is it true that if $\mu$ is a positive bounded regular Borel measure on a bounded interval $[0,2\pi)$ and $f_n:\mathbb{R}\to\mathbb{C}$ are periodic functions satisfying
  $$\int_I f_n(t)d\mu(t)\to 0,\quad\text{as }n\to\infty$$
  then necessarily
  $$\int_I f_n(-t)d\mu(t)\to0,\quad\text{as }n\to\infty?$$

REPLY [7 votes]: Once you have reduced it to the case $\mu \geq 0$, just "evaluate" the integral:
$\int e^{-int}d\mu=\int\cos(nt)d\mu-i\int\sin(nt)d\mu$. So $\hat{\mu}(n) \rightarrow 0$ as $n \rightarrow \infty$ tells you that the real and complex parts of the above integral goes to $0$.
Now, use the fact that $\cos(-nt)=\cos(nt)$ and $\sin(-nt)=-\sin(nt)$ to get your statement.