lebesgue measure basic exercise

Question

I have a basic question about a Lebesgue measure exercise that I am not sure how to solve. (I apologize if this is a simple question, I am new with this subject). Compute the Lebesgue measure of $X$ where $$X =\{\frac{1}{n}; n=1,2,3,...\}.$$ Any help will be very welcome. Cheers

Answer 1

Note that singletons (sets with a single element) in $\mathbb{R}^n$ are null, ie, have measure zero. Therefore, all countable sets are null (specifically, we are looking at a set of the form $X=\{x_n:~n\in\mathbb{N}\}$). This means that since $$X=\{\frac{1}{n}:n=1,2,3,...\}$$ is the countable union of countable sets, it has measure zero with respect to the Lebesgue measure (follows from the fact that a countable set has measure zero).

Answer 2

Every countable subset of $\Bbb R$ has measure $0$. So if $A$ can be expressed as $\{ a_{n} \mid n \in \Bbb N \}$, then $A_{n}$ has measure $0$.

There are two possible arguments to see why. First, remember that if you can write $A$ as a countable union of pairwise disjoint sets, then the measure of $A$ is the sum of the measures of these sets. Well, we have $A = \bigcup \limits_{n = 1}^{\infty} \{ a_{n} \}$, so $m(A) = \sum \limits_{n = 1}^{\infty} m(\{ a_{n} \}) = \sum \limits_{n = 1}^{\infty} 0 = 0$.

Note that in the above, I used the fact that $m(\{a \}) = 0$ for each $a \in \Bbb R$, because $(a - \frac{1}{2^{n}}, a + \frac{1}{2^{n}})$ is a cover of $\{a \}$ for each $n$, and since the measure of $\{a \}$ is the smallest length of open cover of $a$, $0 \leq m(\{a \}) \leq \frac{1}{2^{n - 1}}$ for each $n$, so $m(\{a \}) = 0$.

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Now, for the second possible argument, we cover $\{ a_{n} \mid n \in \Bbb N \}$ with covers that get arbitrarily small, so that the measure of the set will be $0$. Let $B_{m} = \bigcup \limits_{n = 1}^{\infty} (a_{n} - \frac{1}{2^{m}},a_{n} + \frac{1}{2^{m}}) $. Then $B_{m}$ is a cover of $\{a_{n} \mid n \in \Bbb N \}$ for each $m$, and $m(B_{m}) \leq \sum \limits_{n = 1}^{\infty} \frac{1}{2^{m - 1}}$. Clearly, the RHS is finite, and as $m \to \infty$, the $RHS$ goes to $0$, so since $m(A) \leq m(B_{m})$ for all $m$, this implies $m(A) = 0$.