# 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

## 1 Answer

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
• Thanks again! Which teorem allow us to aproximate by step functions? I only know this fact with simple functions instead of step functions. I know step functions aproximation only when a function is riemann integrable. May 9 '13 at 5:47
• @GastónBurrull I don't know any name for it, but it's easy to prove from the approximation by simple functions, so long as you know that every measurable set can be approximated arbitrarily well by a union of finitely many intervals (which is trivial from the definition of outer measure). May 9 '13 at 5:49
• Alex I know the simple aproximation theorem for measurable functions. In this case step function aproximation is valid since function is measurable? or we need to impose integrablility? May 9 '13 at 5:53