How do I show that the integral of $e^{inx}$ over a set of measure $1$ is nonzero for some nonzero $n$? I'm trying to solve a problem from an old qualifying exam that asks to show that $\displaystyle\int_{E}e^{inx}dx$ is nonzero for some integer $n\neq 0$. Here $E$ is assumed to be a measurable set of measure 1.
How do I approach this problem? Do I need to use some result from Fourier analysis or is this a problem from Lebesgue integration theory. So far, the results I know of don't seem to work at all.
 A: Hint: Consider the $2\pi$-periodic function $f=\sum_{n\in\mathbb{Z}} 1_{E+2\pi n}$.  Note that if $g\in L^\infty(\mathbb{R})$ is $2\pi$-periodic, then $\int_E g=\int_0^{2\pi}fg$.  Now what can you say about the Fourier series of $f$, assuming $\int_E e^{inx}dx=0$ for all nonzero integers $n$?
How to finish the proof from here is hidden below:

 If $\int_E e^{inx}dx=0$ for all nonzero $n$, then all of the Fourier coefficients of $f$ are $0$ except the $n=0$ coefficient which is $\frac{1}{2\pi}$ since $E$ has measure $1$.  Since $f\in L^1([0,2\pi))$ and an $L^1$ function is determined by its Fourier series, this means $f(x)=\frac{1}{2\pi}$ almost everywhere.  But this is impossible, since $f$ is integer-valued.

A: For $m\in \mathbb Z,$ let $E_m = E \cap [2m\pi,2(m+1)\pi).$ For $x\in [0,2\pi),$ define
$$\tag 1 g(x) = \sum_{m\in \mathbb Z}\chi_{E_m-2m\pi}(x).$$
Here $E_m-2m\pi$ is $E_m$ translated by $-2m\pi.$ We then have
$$\int_0^{2\pi}g(x)\, dx = \sum_{m\in \mathbb Z}\int_0^{2\pi}\chi_{E_m-2m\pi}(x)\, dx =\sum_{m\in \mathbb Z}\mu(E_m-2m\pi)=\sum_{m\in \mathbb Z}\mu(E_m) =1.$$
We used the MCT to obtain the first equality (permissisble since all functions are nonnegative). So $g\in L^1([0,2\pi),$ and furthermore the sum on the right of $(1)$ converges to $g$ in $L^1.$
Claim: $\int_0^{2\pi}g(x)e^{inx}\, dx\ne 0$ for some integer $n\ne 0.$ The claim will prove the desired result: Why? By the $L^1$ convergence in $(1),$ this integral is is the same as
$$\sum_{m\in \mathbb Z}\int_{E_m-2m\pi}e^{inx}\, dx = \sum_{m\in \mathbb Z}\int_{E_m}e^{in(x-2m\pi)}\, dx = \sum_{m\in \mathbb Z}\int_{E_m}e^{inx} = \int_{E}e^{inx}\,dx.$$
To prove the claim, recall by the uniqueness of Fourier coefficients that the only functions $f\in L^1[0,2\pi)$ with $\hat f (n) = 0$ for all $n\ne 0$ are the constants. So we need only show $g$ is not constant.
There are two cases: a) the sets $E_m-2m\pi$ are "essentially" pairwise disjoint; b) for some pair $j\ne k,m((E_j-2j\pi)\cap (E_k-2k\pi)) > 0.$
In case a), $g$ is a.e. equal to the the characteristic function of a subset of $[0,2\pi)$ having measure $1.$ Since $1 < 2\pi,$ $g$ must then equal $0$ on a set of measure $2\pi -1.$ Hence $g$ is not constant.
In case b), $g\ge 2$ on some subset of $[0,2\pi)$ having positive measure. Obviously $g$ is then $<1$ on a set of positive measure, otherwise $\int_0^{2\pi} g > 2\pi.$ So $g$ is again nonconstant and we're done.
A: I assume your set $E\subset [0,2\pi]$.  From Fourier analysis, we know that linear span of $e^{inx}$ (let it be $A=span\{e^{inx}\}$) is dense in $L^2[0,2\pi]$. Assume the contrary that $\int_{0}^{2\pi}\chi_E(x) e^{inx}dx=0$ for all $n\in \mathbb{Z}$. Then we can pick a sequence $f_n\in A$ such that $f_n\to \chi_E \in L^2[0,2\pi]$. Then by assumption $0=\lim_{n\to \infty}\int_{0}^{2\pi} \chi_E f_ndx=\lim_{n\to \infty}\int_{0}^{2\pi}\chi_E^2dx$ but it is equal to $1$ so we have a contradiction.
