How do I find the lebesgue measure of a interval $[n,n+\frac{1}{n^{2}}]$ when $n\in\mathbb{N}$? I have to use the following definition:

The set-function $\lambda^{n}$ on ($\mathbb{R}^{n}, \mathcal{B}(\mathbb{R}^{n})$) that assigns every half-open $[[a,b)) = [a_{1},b_{1}) \times \dots \times [a_{n},b_{n})\in\mathcal{J}$ the value: $ \lambda^{n}([[a,b))):=\prod_{j=1}^{n}(b_{j}-a_{j}) $ is called n-dimensional Lebesgue measure.

I have considered to write $[n,n+\frac{1}{n^{2}}]$ as a union of distinct half-open intervals, but with no luck. I have tried to write it:

$[n,n+\frac{1}{n^{2}}] = [n,n+\frac{1}{n^{2}}-1) \cup (n+\frac{1}{n^{2}}-1,n+\frac{1}{n^{2}}]$

But then I miss the point $n+\frac{1}{n^{2}}$?

  • 1
    $\begingroup$ Do you know the Lebesgue measure of $[1,2]$? If yes, do you know the Lebesgue measure of $[a,b]$? If yes, do you know the Lebesgue measure of $[a,b)$? $\endgroup$ – Rudy the Reindeer Sep 11 '12 at 9:19
  • $\begingroup$ $(0,1)=\bigcup_{n\in\mathbb N} [2^{-n},2^{1-n})$ implies that $(0,1)$ has measure 1. This is easily adapted to show that $(a,b)$ and $[a,b)$ have same measure. Especially, in 1 dimension, $\{a\}$ has measure 0, hence $[a,b]$ hase the same measure as $[a,b)$. $\endgroup$ – Hagen von Eitzen Sep 11 '12 at 9:26
  • $\begingroup$ I'm a bit confused about both answers so I probably do not even understand the question properly. Given the definition in the OP, why can't we just replace $b$ with $n + \frac{1}{n^2}$ and $a$ with $n$ to get the answer? $\endgroup$ – Rudy the Reindeer Sep 11 '12 at 9:57
  • $\begingroup$ I see. You want a proof that $\lambda [a,b] = \lambda [a,b)$. $\endgroup$ – Rudy the Reindeer Sep 11 '12 at 9:58

This question cannot be answered without assuming some properties of $\lambda$ such as being countably additive, additive, or at least monotone. The last of these properties is the weakest and sufficient to settle the question. For it implies that $$\lambda\big([n,n+1/n^2)\big)\leq\lambda\big([n,n+1/n^2]\big)\leq\lambda\big([n,n+1/n^2+\epsilon)\big)$$ for all $\epsilon>0$. This determines a unique value for $\lambda\big([n,n+1/n^2]\big)$.


Theorem (of continuity of measure)

Let $(X, \mathscr{M}, \mu)$ be a measure space and let $\{A_n\}$ be a sequence of measurable sets.

1) If $A_1 \subset A_2 \subset A_3 \subset \ldots $, then $$\mu\left(\bigcup_{n=1}^\infty A_n\right)=\lim_{n\to \infty} \mu (A_n).$$ 2) If $A_1 \supset A_2 \supset A_3 \supset \ldots$ and $\mu(A_1)<\infty$, then $$\mu\left(\bigcap_{n=1}^\infty A_n\right)=\lim_{n\to \infty} \mu (A_n).$$

You have to apply this. I suggest using 2).


Maybe one could also argue like this:

$[a,b)$ is Lebesgue measurable so by definition, this means that for every set $S \subset \mathbb R^n$ the following holds:

$$ \lambda (S) = \lambda(S \cap [a,b) ) + \lambda ( S \setminus [a,b))$$

In particular, for $S= [a,b]$, we have

$$ \lambda ([a,b]) = \lambda([a,b] \cap [a,b) ) + \lambda ( [a,b] \setminus [a,b)) = \lambda([a,b)) + \lambda (\{b\}) = \lambda([a,b))$$

Now that we know $\lambda ([a,b]) = \lambda([a,b))$ we can apply the definition of $\lambda([a,b))$ to get $$ \lambda [n,n+\frac{1}{n^{2}}] = \lambda ([n,n+\frac{1}{n^{2}})) = n+\frac{1}{n^{2}} - n = \frac{1}{n^{2}}$$


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