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Here the definition of a measure as given in my lecture notes:

A map $ \mu : 2^X \rightarrow [0, \infty] $ is called a measure on X if:

$1. \ \mu(\emptyset) = 0$

$ 2. \ \mu(A) \leq \sum_{i=1}^{\infty} \mu(A_i) $ if $A \subset \bigcup\limits_{i=1}^\infty A_n$

Now, according to my script, 2. implies $\sigma$-subadditivity, i.e.:

$ \mu(\bigcup \limits_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu(A_i) $

My first question is how to explain this implication. My second question is if I understand it correctly that for the case of ifinity many subsets as in 2., the sets don't have to be disjoint. And my last question would be if condition 1. $\mu(\emptyset) = 0$ is not redundant since it follows from the sigma-subadditivity?


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That definition is not correct. – Chris Eagle Feb 23 '13 at 12:06
That's a very strange definition. Why do you want every subset of $X$ to be measurable? – wj32 Feb 23 '13 at 12:19
This definition is from our lecture notes, see page 4: – user62487 Feb 23 '13 at 12:23
The lecture notes use the non-standard terminology of Evans-Gariepy. This means: what is called a measure here is usually called an outer measure. Compare with Bemerkung 1.1.5 on page 4. – Martin Feb 23 '13 at 12:44
Why would anyone downvote this? It's really not user62487's fault if the lecture notes use non-standard terminology and confusing notation. As a beginner in the subject there's really no way to know any better, is there? – Martin Feb 23 '13 at 13:09
up vote 0 down vote accepted

As mentioned in a comment: The lecture notes use the non-standard terminology characteristic of Evans-Gariepy's book Measure theory and fine properties of functions. This means: what is called a measure here is usually called an outer measure. Compare with Struwe's Bemerkung 1.1.5 on page 4. For future questions, you should expect that people will be taken aback by this uncommon usage. I'm not aware of any other introductory text to real analysis and measure theory following Evans and Gariepy's conventions.

For your first question you can simply take $A = \bigcup_{i=1}^\infty A_i$ to conclude $$ \mu(A) = \mu\left(\bigcup_{i=1}^\infty A_i\right) \leq \sum_{i=1}^\infty A_i $$ from condition 2. Notice that in these lecture notes $A \subset B$ really is $A \subseteq B$, meaning that equality $A = B$ is allowed.

Your interpretation is correct: there is no assumption on (pairwise) disjointness of the sets $A_i$.

For your last question, the condition $\mu(\emptyset) = 0$ is not redundant, because $\mu\colon 2^X \to [0,\infty]$ given by $\mu(A) = 1$ for all $A \subseteq X$ satisfies condition 2.

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Great. That was exactly my worry, since I thought A = B is not allowed. By the way, it might also be that this is a definition due to Caratheodory (I have found a source, it is in german though). – user62487 Feb 23 '13 at 13:00
I'm quite sure that the notions of measure, inner measure and outer measure are due to Lebesgue. The definition of a measurable set $E$ via the condition $\mu^\ast(A) = \mu^\ast(A \cap E) + \mu^\ast(A \cap E^c)$ for all $A$ is due to Carathéodory (that's why this is called Carathéodory's condition) while Lebesgue only worked on the Borel subsets of $\mathbb{R}$ IIRC. If you are interested in the history, there are lots of historical notes and excerpts of the original texts in the book Maß- und Integrationstheorie by Elstrodt. – Martin Feb 23 '13 at 13:05
Great, I will have a look. Thank you! – user62487 Feb 23 '13 at 13:07

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