I feel like this should be really easy, but for some reason I don't see how to finish it. I'm probably missing something obvious.(adsbygoogle = window.adsbygoogle || []).push({});

The integral of an integrable simple function ##f=\sum_{k=1}^n a_k\chi_{E_k}## is

$$\int f\,\mathrm{d}\mu=\sum_{k=1}^n a_k\mu(E_k),$$

where the right-hand side is interpreted using the convention ##0\cdot\infty=0##. If we want to use something like this as the definition of the integral, we need to make sure that there's no ambiguity in it. (The problem is that there are many ways to write a given integrable simple function as a linear combination of characteristic functions of measurable sets). One way to do that is to specify the a_k and E_k. We define the a_k by writing the range of f as ##f(X)=\{a_1,\dots,a_n\}## and the E_k by ##E_k=f^{-1}(a_k)##. Now there's no ambiguity in the formula above, so it's safe to use it as a definition.

This raises the question of whether ##f=\sum_{k=1}^n a_k\chi_{E_k}=\sum_{i=1}^m b_i\chi_{F_i}## implies that

$$

\sum_{k=1}^n a_k\mu(E_k)=\sum_{i=1}^m b_i\mu(F_i).

$$ If the E_k are defined as above, it's easy to see that ##\{E_k\}_{k=1}^n## is a partition of X (the underlying set of the measure space). If the ##\{F_i\}_{i=1}^m## is another partition of X, then I find it easy to prove the equality above. However, it holds even when ##\{F_i\}_{i=1}^m## isnota partition of X, right? I first thought that it doesn't, but I failed to find a counterexample, so now I think it does. My problem is that I don't see how to do the proof in the general case.

If ##\{E_k\}_{k=1}^n## and ##\{F_i\}_{i=1}^m## are both partitions of X, then we just write

$$

\begin{align}

& \sum_{k=1}^n a_k\mu(E_k) =\sum_{k=1}^n a_k\mu\bigg(\bigcup_{i=1}^m E_k\cap F_i\bigg) =\sum_{k=1}^n \sum_{i=1}^m a_k \mu\big(E_k\cap F_i\big),\\

& \sum_{i=1}^m b_i\mu(F_i) =\sum_{i=1}^m b_i\mu\bigg(\bigcup_{k=1}^n E_k\cap F_i\bigg)

=\sum_{k=1}^n \sum_{i=1}^m b_i \mu\big(E_k\cap F_i\big),

\end{align}

$$ and then we can easily prove that the left-hand sides are equal by showing that the right-hand sides are equal term for term. Let k,i be arbitrary. If ##E_k\cap F_i=\emptyset##, then ##a_k\mu(E_k\cap F_i)=0=b_i\mu(E_k\cap F_i)##. If ##E_k\cap F_i\neq\emptyset##, then let ##x\in E_k\cap F_i## be arbitrary. We have ##a_k=f(x)=b_i##, and this obviously implies ##a_k\mu(E_k\cap F_i)=b_i\mu(E_k\cap F_i)##.

So...anyone see how to prove or disprove the general case?

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# Simple question about simple functions

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