# Limit of Lebesgue integral in $L_1([-1,1],m)$

Let $([-1,1],\mathcal{M},m)$ be a measurable space in $[-1,1]$ where $m$ is the Lebesgue measure in $\mathbb{R}$ restricted to $[-1,1]$, and $\mathcal{M}$ is the set of $m^*$-measurable subsets of $[-1,1]$ where $m^*$ is the outer Lebesgue measure restricted to $[-1,1]$. If $f\in L_1([-1,1],m)$, show that

$$\lim_{n\to\infty} \int_{-1}^1 f(x)\cos(n x)\, dm=0,$$

Attempt:

LMCT looks not useful since the measurable sequence function $g_n=|f(x)\cos(n x)|$ is not monotone, LDCT looks not useful since $g_n$ can be bounded by $|f(x)|$ but $f(x)\cos(n x)$ probably not converges almost everywhere to any function. I tried to use Fatou's Lemma to show that,

$$\lim_{n\to\infty} \int_{-1}^1 |f(x)\cos(n x)|\, dm=0,$$

and I noted that $\liminf g_n=0$ and Fatou's Lemma apply to $g_n$ says,

$$0\leq \liminf_{n\to\infty} \int_{-1}^1 g_n\, dm$$

which is completelly useless.

• Isn't it Riemann-Lebesgue lemma? May 9 '13 at 5:12
• @julien yes you're right I noted that few minutes ago May 9 '13 at 5:17

For any $\epsilon>0$, we have some step function $g$ such that $\int_{-1}^1|f(x)-g(x)|dx< \epsilon$. Let $g$ take on the values $c_1,\ldots,c_k$ on the intervals $[-1,t_1),[t_1,t_2),\ldots,[t_{k-1},t_k]$ respectively, where $t_0=-1$ and $t_k=1$. Then we have \begin{align} \left|\lim_{n\to\infty} \int_{-1}^1 f(x)\cos(n x)\, dm\right| &\leq \lim_{n\to\infty} \left|\int_{-1}^1 g(x)\cos(n x)\, dm + \int_{-1}^1|(f(x)-g(x))\cos(nx)|\, dm\right|\\ &\leq \lim_{n\to\infty} \left|\int_{-1}^1 g(x)\cos(n x)\, dm + \int_{-1}^1|f(x)-g(x)|\, dm\right|\\ &\leq \lim_{n\to\infty} \left|\int_{-1}^1 g(x)\cos(n x)\, dm\right| + \epsilon\\ &\leq \lim_{n\to\infty} \sum\limits_{i=0}^{k-1}\left|\int_{t_k}^{t_{k+1}}c_k\cos(nx)\, dm\right|+\epsilon\\ &\leq \sum\limits_{i=0}^{k-1}\lim\limits_{n\to\infty} \left|\int_{t_k}^{t_{k+1}}c_k\cos(nx)\, dm\right|+\epsilon\\ &\leq \sum\limits_{i=0}^{k-1}\lim\limits_{n\to\infty} \frac{|c_k|}{n}\left|\int_{nt_k}^{nt_{k+1}}\cos(x)\, dm\right|+\epsilon\\ &\leq \sum\limits_{i=0}^{k-1}\lim\limits_{n\to\infty} \frac{2|c_k|}{n}+\epsilon = \epsilon \end{align} thus letting $\epsilon\to 0$ gives the desired result. The transition from the second to last line to the last is because the integral of $\cos(x)$ over any interval $(a,b)$ has absolute value at most the integral from 0 to $\pi$, which is $2$.
• I didn't get when you replace $nt_{k+1}$ for $\pi$ May 9 '13 at 5:41
• @GastónBurrull The integral $\int_{a}^{b}\cos(x)$, is at most $\int_0^\pi\cos(x)$, since $\cos(x)$ is periodic with period $2\pi$ and the integral over its period is $0$. May 9 '13 at 5:44