If $-\infty <a<b<\infty$ and $\mu$ is $1/(b-a)$ times Lebesgue measure on $[a,b]$, then $\mu([a,b])=1$

Question

If $-\infty <a<b<\infty$ and $\mu$ is $1/(b-a)$ times Lebesgue measure on $[a,b]$, then $\mu([a,b])=1$.

What's the meaning of this sentence? What is $1/(b-a)$ times Lebesgue measure on $[a,b]$?

Answer 1

It is just the measure $$\mu(A) = \frac{1}{b-a} \lambda(A), \quad\forall A \in \mathcal{B}(\mathbb{R})$$

where $\lambda$ is the Lebesgue measure.

Indeed then

$$\mu([a,b]) = \frac{1}{b-a}\lambda([a,b]) = \frac{1}{b-a} \cdot (b-a) = 1$$

Answer 2

$\mu(A)=\dfrac{1}{b-a}|A|$, where $|\cdot|$ is the usual Lebesgue measure, then $\mu([a,b])=\dfrac{1}{b-a}|[a,b]|=\dfrac{1}{b-a}\cdot(b-a)=1$. In a sense, $\mu$ is the normalized Lebesgue measure.