Proving $\lambda((a,b))= \lambda([a,b])=\lambda((a,b])=b-a$

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How to prove that $\lambda((a,b))= \lambda([a,b])=\lambda((a,b])=b-a$, where $\lambda$ is the Lebesgue measure?

It seems intuitive, but I'm confused as I've seen so many different definitions of the Lebesgue measure.

The definition that I have been given is just that $\lambda$ is a measure $\mathcal B(\mathbb R)\to[0,\infty]$ so that for any $a,b\in\mathbb R, a<b: \;\lambda((a,b])=b-a$. I have not learned about outer measures.

How am I supposed to apply this to intervals that are not in the form of $(a,b]$? Also, how to apply this when $a = -\infty$ or $b=\infty$?

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Let $\frac{b-a}{2}>\epsilon>0$ . Then $(a+\epsilon,b] \subset [a,b] \subset (a-\epsilon,b+\epsilon]$

Thus $b-a-\epsilon \leq \lambda([a,b]) \leq b-a+2\epsilon$

Thus sending $\epsilon \to 0$ you have the conclusion.

Use this idea for other forms of bounded intervals.

Now for an interval $I=(a,+\infty)$ we have that $I \supset (a,a+n],\forall n \in \Bbb{N}$ thus $\lambda (I) \geq n, \forall n \in \Bbb{N}$

So $\lambda (I)=+\infty$.

Now use these two ideas in my answer to finish your problem for the other forms of unbounded intervals.