Lebesgue measure 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 $n+\frac{1}{n^{2}}$?
 A: 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 $\lambda\big([n,n+1/n^2]\big)$. 
A: 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).
A: 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
$$ \lambda [n,n+\frac{1}{n^{2}}] = \lambda ([n,n+\frac{1}{n^{2}})) = n+\frac{1}{n^{2}} - n = \frac{1}{n^{2}}$$
