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Let $\Omega=(0,1]$ with Borel field and $A=\bigcup_{i=1}^I(a_i,b_i]$ such that $0\le a_1\le b_1\le a_2 \le b_2 \le ...\le b_I$ $$\lambda(A) = \sum_{i=1}^I(b_i-a_i)$$ Then, how do we show that $\lambda$ is well-defined?

My attempt is that let $A$ and $A'$ be an arbitrary sets in Borel field such that $A=A'$, then I need to show $\lambda(A)=\lambda(A')$. I have tried to prove by contradiction, but I could not do it...

Is there other approach to solve this problem?

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  • $\begingroup$ I think you've very much misunderstood the definition. $\lambda(A)$ is not defined for arbitrary sets, but only for things that are finite unions of such half-open intervals - you can't just assume that $A$ and $A'$ are arbitrary sets. $\endgroup$ – user296602 Aug 24 '16 at 21:37
  • $\begingroup$ @T.Bongers then how do we show that $\lambda(A) = y,\lambda(A')=y'$, then $y=y'$, where $A=A'$, when we have infinitely many representation of A? $\endgroup$ – user1292919 Aug 24 '16 at 21:54
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    $\begingroup$ You're supposed to show that $\lambda$ agrees for any two different representations of $A$: $A=\cup_{i=1}^I(a_i,b_i]=\cup_{i=1}^{I'}(a'_i,b'_i]$ $\endgroup$ – Alex R. Aug 24 '16 at 22:06
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The $\lambda$ you defined is clearly well-defined (as long as $I < \infty$). The difficulty is to prove that you can build a measure that extend your $\lambda$. The Lebesgue measure (on the real line) commonly means a (the) Borel measure that coincides with the length on intervals.

If your goal is to prove that the Lebesgue measure exists and has nice properties a fast way to do it is to consider the sequence of probability measures on $[0,1]$, $$\mu_{n} = \sum_{i=1}^{n} \frac{1}{n} \delta_{\frac{i}{n}}$$ and show (for instance with the Prokhorov theorem) that $\mu_{n}$ converges to a limit probability measure that you call $\lambda_{[0,1]}$, the Lebesgue measure on $[0,1]$. You can repeat this procedure on any interval $[n,n+1]$ ($n \in \mathbb{Z}$) and you obtain the Lebesgue measure on $\mathbb{R}$ by gluing them all together: $$\lambda = \sum_{n \in \mathbb{Z}} \lambda_{[n,n+1]}.$$ You immediately obtain that for a Riemann integrable function $f$, $$\int f(x) dx = \int f d \lambda$$ which implies the property that $\lambda([a, b]) = b - a$.

If you want to use an approach similar to the one in you initial question then you introduces an outer measure $$\lambda^{*}(A) = \inf \{ \sum_{i \in \mathbb{N}} (b_{i} - a_{i}) \, : \, A \subset \bigcup_{i\in \mathbb{N}} ]a_{i}, b_{i}[ \},$$ where the infimum is over all countable covering of $A$ by open intervals $]a_{i}, b_{i}[$. Then you show that the space of $\lambda^{*}$-measurable sets contains the Borel $\sigma$-field. By a general theorem the restriction of $\lambda^{*}$ to its measurable sets is a measure and therefore the restriction to the Borel sets (that you call $\lambda$) is also a measure. Then you compute by hand that $$\lambda^{*}(]a,b[) = \lambda^{*}([a,b]) = b -a.$$

From my point of view, the second approach is heavier. You should note that other approaches exist such as the one in Rudin using Riesz duality theorem.

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