Cantor-Lebesgue's theorem

Question

I have this proof of the Cantor-Lebesgue theorem that was given to me by my professor, but I don't understand a step of it.

The statement is:
Let $E\subset\Bbb{R}$ , measurable, $\infty>m(E)>0$.
Let $ f(x) =$ $\sum_{i=1}^{\infty}$ $a_n\cos(nx) + b_n\sin(nx)$ with $a_n,b_n$ $\in$$\Bbb{R}$ and we suppose that $\forall x\in E$, the series which define $f(x)$ converge.
Then $a_n$, $b_n$ $\to$ 0 for $n\to +\infty$.

$Proof$:
Let $f_n(x)= a_n\cos(nx) + b_n\sin(nx)$.
Firstly, we see that exist $r_n$, $\theta_n$ such that $a_n = r_n\cos(\theta_n)$, $b_n=r_n\sin(\theta_n)$, in fact if $a_n=0$ and $b_n=0$, $r_n=0$ and $\theta_n=0$ are good.
If $a_n\neq 0$ or $b_n\neq 0$, we can write:
$$a_n =\sqrt{a_n^2 \,+\, b_n^2 }\cdot\frac{a_n}{\sqrt{a_n^2 \,+\, b_n^2 }},\quad b_n =\sqrt{a_n^2 \,+\, b_n^2 }\cdot\frac{b_n}{\sqrt{a_n^2 \,+\, b_n^2 }}$$ so $r_n$ = $\sqrt{a_n^2 \,+\, b_n^2 }$, $\,$ $\cos(\theta_n)$ = $\frac{a_n}{\sqrt{a_n^2 \,+\, b_n^2 }}$, $\,$ $\sin(\theta_n)$ = $\frac{b_n}{\sqrt{a_n^2 \,+\, b_n^2 }}$
(it's well known that with this definition such $\theta_n$ exist).

Now, if we substitute in the definition of $f_n$ and we apply the trigonometric formula of subtraction of angles for the $\cos(x)$ function, we obtain:

$$f_n(x)=r_n\cdot\cos(nx-\theta_n)$$

Now, given that $f(x)$ converge for all $x$ in $E$, then the principal term of the series ( $f_n(x)$ ) vanished for $n$ that tends to +$\infty$.

If $a_n$ or $b_n$ don't go to zero, then $r_n$ don't go to zero, then necessarily $\cos(nx-\theta_n)$ goes to zero $\forall x \in E$, and so even $\cos^2(nx-\theta_n)$ goes to $0$.

Now, $\cos^2(nx-\theta_n) \to 0$, $|\cos^2(nx-\theta_n)|I_E(x) \le I_E(x)$ ( $I_E(x)$ is the indicator function of E), and $I_E(x) \in L^1(E)$ imply, by dominated convergence, that $$\lim_{n\to\infty} \int_{E} \cos^2(nx-\theta_n) = 0$$

We can write also $\forall n$: $$\int_{E} \cos^2(nx-\theta_n) = \int_{E} {1\over 2} + \int_{E} {\cos(2nx-2\theta_n)\over2}$$

Now, if we take the limit in $n$, we obtain $0 = {m(E) \over 2}$, and this is wrong because $m(E)>0$.

I have written this in my notes, but I don't understand why the second integral at the RHS vanished, because ${\cos(2nx-2\theta_n)\over2}$ should converge to $-{1\over2}$.

Maybe have I misunderstood what the professor said, or can this proof be saved?

Thanks to everyone will read this.

Answer 1

I don't know why you think ${cos(2nx-2\theta_n)\over2}$ should converge to $-{1\over2}$. For typical $x$ that quantity oscillates as $n\to\infty$.

Write $$cos(2nx-2\theta_n)=\cos(2nx)\cos(2\theta_n)+\sin(2nx)\sin(2\theta_n).$$

Since the sine and cosine of $2\theta_n$ are bounded the Riemann-Lebesgue Lemma (applied to $\chi_E$) shows that the integral tends to $0$ as $n\to\infty$.

Answer 2

Here is another version of the Cantor-Lebesgue theorem that is a little more general than the one stated in the OP:

Theorem (Cantor-Lebesgue) Let $(c_n:n\in\mathbb{Z})\subset\mathbb{C}$ be a sequence such that the trigonometric series \begin{align}\sum_{n\in\mathbb{Z}} c_n e^{int}\tag{0}\label{zero}\end{align} converges on a set of positive Lebesgue measure. Then \begin{align}\lim_{|n|\rightarrow\infty}c_n=0\tag{1}\label{TS}\end{align}

Before presenting a proof of this Theorem we make a few remarks:

1. Convergence of a series $\sum_{n\in\mathbb{Z}}a_n$ means $\lim_\limits{m\rightarrow\infty}\sum_{|k|\leq m}a_m$ converges.
2. The trigonometric series $\frac12a_0+\sum^\infty_{n=1}a_n\cos nt + b_n\sin nt$ with $a_n,b_n\in\mathbb{R}$ is of the form \eqref{zero}: Indeed, from \begin{align} e^{ix}=\cos x + i\sin x \end{align} we have that $$ \sum^\infty_{n=1}a_n\cos nt + b_n\sin nt=\sum_{n\in\mathbb{Z}\setminus\{0\}}c_n e^{int}=\sum^\infty_{n=1}\big(c_{-n}e^{-int}+c_ne^{int}\big) $$ where $c_{-n}=\overline{c_n}$, and $c_n=\frac12(a_n-ib_n)$.

Proof of Cantor-Lebesgue's theorem:

The result will follow from the following result:

Lemma: Let $(c_n:n\in\mathbb{Z})\subset\mathbb{C}$ is a sequence such that the sequence of functions defined as $A_n(t):=c_{-n}e^{-int}+c_ne^{int}$ converges to $0$ on set of positive Lebesgue measure. Then $\lim_{n\rightarrow\infty}\sqrt{|c_n|^2+|c_{-n}|^2}=0$.

Proof:

Suppose $A_n(t)\xrightarrow{n\rightarrow\infty}0$ for all $t\in F\subset[-\pi,\pi]$ where $|F|>0$. For each $n\in\mathbb{Z}_+$ define $r_n:=\sqrt{|c_n|^2+|c_{-n}|^2}$.

Suppose $\{r_n:n\in\mathbb{Z}_+\}$ is bounded. Then, by dominated convergence \begin{align} \lim_{n\rightarrow\infty}\int_F |A_n(t)|^2\,dt=0\tag{2}\label{two} \end{align} From $$|A_n(t)|^2=|c_{-n}|^2+|c_n|^2+c_{-n}\overline{c_n}e^{-2int} + c_{n}\overline{c_{-n}}e^{2int},$$ the boundedness of the $\{c_n:n\in\mathbb{Z}\}$, and an application of the Riemann-Lebesgue's lemma we obtain that \begin{align} c_{-n}\overline{c_n}\int_F e^{-2int}\,dt+c_{n}\overline{c_{-n}}\int_F e^{2int}\,dt\xrightarrow{n\rightarrow\infty}0. \end{align} Consequently, $$0=\lim_n\int_F|A_n(t)|^2\,dt=|F|\lim_n r^2_n$$ Hence $r_n\xrightarrow{n\rightarrow\infty}0$.

If $\{r_n:n\in\mathbb{Z}_+\}$ is not bounded, then $r_n$ does not converge to $0$ as $n\rightarrow\infty$. Hence, for some $\varepsilon>0$ and some subsequence $n_k$, $r_{n_k}\geq\varepsilon$. Define e $B_m(t)=\frac{1}{r_m}A_m(t)$ when $m=n_k$, and $B_n(t)=0$ otherwise. Then, the sequence $B_m$ is of the form $$B_m(t)=c'_{-m}e^{-imt}+c_me^{imt}$$ where $r'_{m}=\sqrt{|c'_{-m}|^2+|c'_m|^2}$ forms a bounded sequence. Also, $B_m$ converges to $0$ on $F$. From case (1) it follows that $r'_m\xrightarrow{m\rightarrow\infty}0$, this is in contradiction to the fact that $\limsup_mr'_m=1$. Hence $\{r_n:n\in\mathbb{Z}_+\}$ must be bounded. The conclusion of the Lemma follows.