Find the length of a finite interval

Question

Let $\mathcal{I}=\{[a,b)| a,b \in \mathbb{R}, a<b\}$

Define the length function $\ell(I)=b-a$ where $I \in \mathcal{I}$

The question asks to deduce that the length of an open interval $(a,b) = b-a$ given the above definitions

-Attempt-

since, $(a,b)=\lim_{n \to \infty} [a - \frac{1}{n},b)$ where $n \in \mathbb{N}^*$

Now $\ell([a-\frac{1}{n},b))=b-a+ \frac{1}{n}$ so by the continuity of $\ell$ we get $\lim_{ n \to \infty} \ell([a-\frac{1}{n},b))=\ell(\lim_{ n \to \infty}([a-\frac{1}{n},b)))= b-a$

I am not sure of it really. Any thoughts?

EDIT: The properties of the length function $\ell$ are the usual well-known properties of the length function on real intervals. ($\ell$ is non-negative, $\ell(\phi)=0$, monotone, countably additive and sub-additive)

Answer 1

Set $\delta = (b-a)/2$. If you have countable additivity, you should be able to write something like $$ (a,b) = \big(\bigcup_{k=0}^\infty [a+\delta/2^{k+1},a+\delta/2^k)\big) \cup [a+\delta,b) , $$ and then use that $\delta/2^k-\delta/2^{k+1}=\delta/2^{k+1}$ and $\delta \sum_{k=0}^\infty 1/2^{k+1} = \delta \sum_{k=1}^\infty 1/2^k = \delta$.

Answer 2

If we want a continuous continuation of $\ell$ to open intervals, then the limit and the function must commute and $$\lim_{\epsilon\to0^+}\ell([a+\epsilon,b))=\ell(\lim_{\epsilon\to0^+}[a+\epsilon,b))=\ell((a,b))$$ while obviously

$$\lim_{\epsilon\to0^+}\ell([a+\epsilon,b))=b-a.$$

---

If you don't like the shortcut

$$\lim_{\epsilon\to0^+}[a+\epsilon,b)=(a,b),$$ you can as well require the monotonicity of the continuation and use $(a,b)=\bigcup_n [a_n,b)$ where $a_n$ tends to $a^+$ and deduce $\sup_{a_n\to a^+}\ell([a_n,b))$ for a definition.