How do I show that the integral of $e^{inx}$ over a set of measure $1$ is nonzero for some nonzero $n$?

Question

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.

Answer 1

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.

Answer 2

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.