I am having trouble understanding why the following two statements are equivalent:
First we denote by $A$ the disc algebra consisting of complex continuous functions in the unit circle such that $\int_{-\pi}^{\pi}f(\theta)e^{in \theta}d\theta =0$ for all $n>0$
(Here $f$ is defined by identifying the continuous function on the circle with a continuous function on [-$\pi,\pi$].)
1) I want to prove that the real parts of functions in $A$ are uniformly dense in the space of real continuous functions on the circle and my book says this equivalent to proving the following:
2) If $\mu$ is a finite real baire measure on the unit circle ( $:=T$) such that $\int_Tfd\mu=0$ for all $f\in A$ then $\mu$ is the zero measure.
I don't understand why proving the second statement implies the first.
Any help will be greatly appreciated.
You say you want to show (2) implies (1). The most reasonable interpretation is that you want to show (2) implies (1) for any collection $A$ of continuous, complex function on $T$ (if you were only speaking about the $A$ you defined at first, you can just prove (1) and/or (2)). So I will speak with $A$ in that level of generality.
It is obvious then that (2) does not imply (1). Indeed, let $A = \{ip(x) : x \in T\}$ for all polynomials $p \in \mathbb{R}[x]$. Then $\int_T fd\mu = 0$ for all $f \in A$ means $\int_T p(x)d\mu = 0$ for all $f \in A$, which means $\int_T f(x)d\mu = 0$ for any $f \in C(T)$, and so $\mu = 0$. However, $Re[ip(x)] = 0$ for all $ip(x) \in A$ and so the real parts of functions in $A$ are not at all uniformly dense in the set of real continuous functions on $T$.
The implication that you would want to prove is that (1) is implied by the variant of (2):
(2') If $\mu$ is a finite Baire measure on $T$ such that $\int_T Re[f]d\mu = 0$ for each $f \in A$, then $\mu \equiv 0$.
Here is a proof that (2') implies (1): Let $\tilde{A} = \{Re[f] : f \in A\}$. Suppose $\tilde{A}$ is not dense in $C(T)$. Then there is an element of $C(T)^*$ that is $0$ on $\tilde{A}$ but not identically $0$; as $C(T)^* = M(T)$, the finite Borel measures on $T$, we get a measure $\mu \not \equiv 0$ with $\int_T Re[f]d\mu = 0$ for all $f \in A$.