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I am trying to understand the proof of the following theorem:

**There exists only one linear application $Int: \epsilon (\mathbb{R^n}) \rightarrow \mathbb{R}$ called integral s.t $Int(\chi_P) = \mu(P)$ **

First, let's define some of the terms:

  • $P$ is a subset of $\mathbb{R^n}$. $P = I_1 \times I_2 \times ... \times I_n$ with $I_i$s intervalls

  • $\epsilon (\mathbb{R^n}) $ is the vector space of simple functions

  • $\mu(P)$ is the measure of $P$

Understanding the proof

How they prove that theorem is as follows.

Let us say $f = \Sigma_{i=1}^{k} c_i \chi_{P_i}$

We therefore have $Int(f) = \Sigma_{i=1}^{k} c_i \mu(P_i)$

Now they suppose that f(x) = 0 and they say that all you have to do is show that $int(f) = 0$ I don't understand how that is a general proof. and also how does it show unicity?

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  • $\begingroup$ Doesn't the supposition that $f$ is linear guarantee that $f(0)=0$? I don't see why this is something that needs proving. $\endgroup$
    – aduh
    Jan 4 '17 at 16:00
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I don't know how $f=0$ implies $\text{Int} f = 0$ shows uniqueness.

In any case, it follows in a straightforward way from linearity and the measure condition.

Suppose $I$ is another linear functional satisfying the condition, then $I(\sum_k c_k 1_{P_k}) = \sum_k c_k \mu P_k = \text{Int} (\sum_k c_k 1_{P_k})$ from which we get $I=\text{Int}$.

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