# Compute integral of a Lebesgue measurable set

Let $B \subset [0,2\pi]$ be a Lebesgue measurable set. Prove that:

$\displaystyle \lim_{n \to \infty} \int_{B} \cos(nx) dx = 0$

OK I did this assuming B is an open interval, this is pretty easy using the fact that the sine function is bounded by 1. Now I'm stuck in the general case, I'm somewhat confused by "Lebesgue measurable" I know this means that the measure is given by the outer measure i.e the infimum of the sum of all measures of open covers of B. But I'm having trouble writing it, I get confused when working with Lebesgue measurable sets. Can you please help me?

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@Marc: Well, for the Lebesgue integral to even be defined you need the domain of integration to be Lebesgue measurable... – Arturo Magidin Nov 3 '10 at 19:20
Can I proceed like this? if B = (a,b) then using the FTC we can easily see the limit is zero. Now assume B is the disjoint union of open intervals, so the integral over B is equal: $\sum_{j=1}^{\infty} \int_{(a_{j},b_{j})} f$ and then? – student Nov 3 '10 at 19:26

## 3 Answers

hint

Re-write the integral as the following

$$\int_0^{2\pi} \chi_B(x) \cos(nx) dx$$

where $\chi_B$ is the characteristic function of the set $B$ (so that it equals 1 on $B$ and 0 else where).

Now, you've already proven the case where $B$ is an open interval. Now take a sequence of decreasing coverings for $B$ by finitely many disjoint open intervals. For each covering, the result is true by what you've already shown. Take the limit using dominated convergence theorem (if $B \subset C$, then $|\chi_B \cos| \leq |\chi_C \cos|$).

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Each Lebesgue integrable function can be approximated by a finite step function in norm. That is if $f\in L^1$ and $\varepsilon>0$ then there are intervals $I_1,\ldots,I_N$, and scalars $a_1,\ldots,a_N$ such that $$\int |f-\sum_{n=1}^N a_n\chi_n|<\varepsilon.$$ Where $\chi_n$ is the characteristic function on $I_n$, that is $\chi_n(x)=1$ for $x\in I_n$ and $\chi_n(x)=0$ otherwise.

Now, on each bounded interval $I$ you have already proved that $\int_I \cos(nx)dx\to 0$, hence given $\varepsilon>0$ there is a step function as above and we get $$\limsup\left|\int_B\cos(kx)dx\right|= \limsup\left|\int (\chi_B-\sum_{n=1}^N a_n\chi_n +\sum_{n=1}^N a_n\chi_n)\cos(kx)dx\right|$$ $$\leq \limsup\int |\chi_B-\sum_{n=1}^N a_n\chi_n|dx +\limsup\left|\int\sum_{n=1}^N a_n\chi_n\cos(kx)dx\right|$$ $$\le\varepsilon + \sum_{n=1}^N |a_n|\cdot \limsup\left|\int\chi_n\cos(kx)dx\right| =\varepsilon +0.$$ From which you conclude the result since this holds for any $\varepsilon>0$.

This works not only for $\chi_B$, but for any $f\in L^1$ - it is called the Riemann-Lebesgue Lemma.

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Thanks to both. I'm still confused because books define Lebesgue measurable sets in different ways (Caratheodory extension, outer measure, etc). What is the exact definition you are using for Lebesgue measurable set? I'm sure once I understand how it is defined I will be able to understand fully your hint(s). Thanks again.

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Which book do you read? There are several settings leading to the very same Lebesgue integral. – AD. Nov 3 '10 at 20:25
Bartle, Folland, Zygmund, etc. Here is my try: taken an open cover of B consisting of countable disjoint intervals (I don't get why it must be finite so I take it countable) of open intervals. Define $g_{n}(x)= \chi{\cup_{k=1}^{n} (a_k,b_k)} cos(nx)$ then g_n converges pointwise to $\chi_{B}(x)cos(nx)$, g_n is increasing so we can use monotone convergence theorem to take out the series and then integrate term by term but each of the remaining integrals are integrals over open intervals which we already know they are 0. OK? – student Nov 3 '10 at 20:39
@Marc: sounds about right; I chose that at each step you approximate by finitely many because a sum of finitely many $\epsilon$s is still small. The total number of intervals used increases for each subsequent approximation. So no, $B$ is not necessarily covered by finitely many intervals, but it is approached by a sequence of covers, each having finitely many intervals, but the number increases. Sorry for the confusion. – Willie Wong♦ Nov 3 '10 at 21:09
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