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Let $u_n$ denote the uniform distribution on $(n,n+1]$. We define the Lebesgue Measure $m: \mathbb{B} \to \mathbb{R}^{+}$ by $$ m(A) = \sum_{n \in \mathbb{Z}} u_n(A)$$

We were given this definition of the Lesbesgue measure in class but I am struggling a little bit with it

My thoughts

I wonder why it is not defined by

$$m(A) = \sum_{n \in \mathbb{Z}} u_n(A \cap (n,n+1]) $$

since by the definition of the uniform distribution on $(n,n+1]$ I know that its density is $$ \mathbb{I}_{(n,n+1]}(A)$$ or in other words

$$ f(A) =\begin{cases}1 &\text{A $\in$ (n,n+1]}\\0&\text{else}\end{cases} $$

However, if you choose any set $A=(a,b]$, since on of the desired properties of the Lebesgue measure is $m((a,b])=b-a$, that is "bigger" than $(n,n+1]$ you will never get a measure for it because $(a,b]$ won't be in $(n,n+1]$.

So me question is does this definition only hold if also $A$ is chosen to be between some $(n,n+1]$, which does not make sense for me, or am I missing an important point or is there an issue with the definition.

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    $u_n$ should be a (probability) measure. However, later on you write something like "measure=function" if I understand your notations correctly. The last $u_n(A)=1$ or $0$ makes no sense to me since $A$ is a set. – 2017-01-15
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    $u_n$ should be the uniform distribution on $(n,n+1]$ of which the density is given - as far as I understood - as it an indicator variable $I_{(n,n+1]}(A)$ – 2017-01-15
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    I edited it - is it now clear? – 2017-01-15
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    All is as you say, but it is a quite inappropriate way to introduce "Lebesgue measure". Should we say that $u_n(A)$ is a Lebesgue integral with respect to an undefined measure...? – 2017-01-15

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