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Proposition: Let $f$ be a bounded measurable function on a set of finite measure $E$. Suppose $A$ and $B$ are disjoint measurable subsets of $E$. Then $$\int_{A \cup B} f =\int_A f + \int_B f.$$ Furthermore, let $E_0$ be any measurable subset of $E$. Then $\int_{E_0} f = \int_E f\cdot \chi_{E_0}$.

The proof of the above proposition relies on the proof of $\int_{E_0} f = \int_E f\cdot \chi_{E_0}$ which was left as an exercise. Intuitively the formula makes sense. Proving it on the other hand is where I'm a bit stuck. Since $f$ is bounded I thought about defining two sets of simple functions $\phi_{n,E}, \phi_{n,E_0}, \psi_{n,E}, \psi_{n,E_0}$ and then applying the definition.

Is there a more efficient way of proving this assertion: $$\int_{E_0} f = \int_E f\cdot \chi_{E_0}$$

The above proposition in Royden requires this excercised to be proven in order to fully understand (or appreciate) the proof of the proposition:

Suppose $f$ is a bounded measurable function on a set of finite measure $E$. Let $E_0 \subseteq E$. Then $\int_{E_0} f = \int_E f \cdot \chi_{E_0}$.

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2 Answers 2

up vote 1 down vote accepted

It depends on how $\int_{E_0} f$ has been defined. I don't have Royden on hand, but if he hasn't defined it by $\int_{E_0} f = \int_E f \cdot \chi_{E_0}$, a possible alternative definition could be phrased as follows:

Let $(E, \Sigma, \mu)$ be a measure space, and let $E_0 \in \Sigma$ be some measurable subset. We define the measure space $(E_0, \Sigma_0, \mu_0)$ by $A_0 \in \Sigma_0 \iff A_0 = A\cap E_0$, for some $A \in \Sigma$. Note that $\Sigma_0 \subset \Sigma$, so we can simply take $\mu_0$ to be the restriction of $\mu$ to $\Sigma_0$.

Now we have a definition for $\int_{E_0} f d\mu_0$, for nonnegative measurable f as $\sup\{\int_{E_0} s(x) d\mu_0\}$, where the supremum is taken over the integral of all nonnegative simple functions on $E_0$ satisfying $s \le f$.

We'd like to see that, using this definition, $\int_{E_0} f d\mu_0 = \int_E f \cdot \chi_{E_0} d\mu$. To prove this, first prove it under the assumption that $f$ is simple and show that the result follows in general from this.

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I think he is asking us to prove the definition. I also don't have a copy of Royden in front of me. But in his proof of the above proposition he makes the remark that for any measurable set $E_0 \subseteq E$ we have $\int_E f\cdot \chi_{E_0} = \int_{E_0}f$. –  emka Nov 27 '12 at 3:13
    
You cannot prove a definition. In all likelihood, what I have outlined above is what's intended, as it's something worth remarking on anyway since you have two competing definitions for the integral (one by considering $\int_E f \cdot \chi_{E_0}$ and the other by considering $E_0$ as a measure space in its own right) and it's not a priori obvious that they should agree. However, as noted in the answer above mine, that equality is often taken as the definition of integrating over a subspace. –  anonymous Nov 27 '12 at 3:46

Usually, the integral over $E_0$ is defined to be $\int_E f\cdot\chi_{E_0}^{\phantom{E}}$.

What would be the definition of $\int_{E_0}f$ otherwise?

The only thing that comes to mind is to define $\int_{E_0}f=\int_Ef|_{E_0}$. In that case, all you need to show is that $f|_{E_0}=f\cdot\chi_{E_0}^{\phantom{E}}$.

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It makes sense. I added something to my question if that may help. –  emka Nov 26 '12 at 0:49
    
It's also possible to define $\int_{E_0}f\,d\mu$ as $\int f|_{E_0}\,d\mu_{E_0}$ where $\mu_{E_0}$ is the restriction of $\mu$ to $E_0$, i.e. $\mu_{E_0}(A) = \mu(A \cap E_0)$ –  kahen Nov 26 '12 at 0:57
    
@kahen: I agree, although I fail to see what is gained from such definition, beyond having to spend time proving things that are otherwise obvious. –  Martin Argerami Nov 26 '12 at 1:01

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