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Let $\Delta_1$ be the set {[a, b]; $a \leq b, a, b \in \mathbb{R}$}.

Let $\Delta_2$ be the set {[a, b); $a \leq b, a, b \in \mathbb{R}$}.

Let $\Delta_3$ be the set {(a, b]; $a \leq b, a, b \in \mathbb{R}$}.

Let $\Delta_4$ be the set {(a, b); $a \leq b, a, b \in \mathbb{R}$}.

Let $\Delta = \Delta_1 \cup \Delta_2 \cup \Delta_3 \cup \Delta_4$.

Let $I \in \Delta$. If $I$ is one of the forms [a, b], [a, b), (a, b], (a, b), We define $\mu(I)$ = b - a.

Let $\Delta^n$ be the set {$I_1 \times ... \times I_n: I_1, ..., I_n \in \Delta$}.

The following proposition is trivial if we assume the theory of Lebesgue measure. However, I think it'd be nice if we could prove it by an elementary method.

Proposition There exists a unique function $\mu^n: \Delta^n \rightarrow \mathbb{R}$ which satisfies the following properties.

(1) $\mu^n(I_1 \times ... \times I_n) = \mu(I_1) \times ... \times \mu(I_n)$ for $I_1, ..., I_n \in \Delta$.

(2) Let $A, A_1、...、A_m \in \Delta^n$. Suppose $A$ is a disjoint union of $A_1、...、A_m$. Then $\mu^n(A) = \mu^n(A_1) + ... + \mu^n(A_m)$

EDIT The above proposition can be generalized using the following notion. Let $X$ be a set. Let $\Gamma$ be a subset of the power set $P(X)$. We call $\Gamma$ a semiring on $X$.

(1) $\emptyset \in \Gamma$.

(2) If $A, B \in \Gamma$, then $A \cap B \in \Gamma$.

(3) If $A, B \in \Gamma$, then $A - B$ is a disjoint union of finitely many elements of $\Gamma$.

The above $\Delta$ is a semiring on $\mathbb{R}$.

EDIT By this result, it suffices to prove the case n = 1 which is almost trivial.

EDIT I came up with this problem as an application of the notion of semiring and this. I found the notion useful.

EDIT A nice little feature of this proposition as a fundamental lemma of Lebesgue measure theory is that it is symmetric, i.e. you don't have to use only intervals of type [a, b)(or (a, b]).

EDIT Since some people seem to think this might be my homework or a textbook exercise, let me write my opinion for posting homework/exercise as a question. Firstly I'm not a student so that I don't have, nor will have any homework. Nor this is a textbook exercise(I came up with it myself). Secondly when I answer a question, I don't care whether it is homework/exercise or not. I also don't care whether the questioner shows effort or not. If I think it's uninteresting or too difficult, I won't answer. Otherwise, and if I can solve it, I'll post an answer. Usually I try to write a full proof. If the questioner thinks that reading a full proof is not good for himself, he can always skip a part of it. In short, I solve a problem as a problem. I think this strategy makes things simpler and works well most of the time, though it might not be very pedagogical.

EDIT Generally speaking, IMO, a mathematical problem should be valued only by its inherent value(not that I'm implying my question is of great value, though). What matters is whether it is interesting and/or useful or not. Whether it is homework/exercise or not, and whether the questioner pays enough effort to solve it or not, do not matter, or are of secondary importance. Interesting and/or useful problems are posted and people post the answers. This is the best part of this forum, IMO.

EDIT Anyway, I wrote what I had done so far as GEdgar suggested.

EDIT May I ask the reason for the downvote so that I could improve my question? I think the question is clear. I wrote what I had done so far. I also wrote my motivation. I appreciate if you'd kindly inform me what's wrong with my question.

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How is $\mu^n$ defined for something which is not a product of intervals? – Davide Giraudo Jun 22 '12 at 21:27
I editted my question. – Makoto Kato Jun 22 '12 at 21:34
I editted again. Thanks. – Makoto Kato Jun 22 '12 at 21:49
Well, equation (1) defines $\mu^n$ on the set $\Delta^n$. So much for existence and uniqueness. The proof of (2) is easy but boring: partition $A$ by all hyperplanes $x_i=a_i$ and $x_i=b_i$, write $\mu^n(A)$ as product of sums, multiply out, collect the terms into $\mu^n(A_i)$ – user31373 Jun 22 '12 at 22:37
This problem is in the text of Kolmogorov & Fomin. (Whether from the original, or one of the problems added by Silverman for the English translation, I don't know.) The solution is as suggested by Leonid. – GEdgar Jun 24 '12 at 2:59

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