$\mu^*$ measurable sets

A subset E of X is said to be $\mu^*$ measurable if

$\mu^*(A) = \mu^* (A \cap E) + \mu^*(A \backslash E)$ for all subsets A of X.

In other words what this says is a set E is $\mu^*$ measurable if E and its compliment are sufficiently separated that they divide an arbitrary set A additively.

What is the intuition behind this definition? What is this saying?

I know $\mu^*$ means outer measure but what does the term $\mu^*$ measurable mean?

Does it just mean that the outer measure of a subset can be found?

• Sets with that property are by definition $\mu^{*}$-measurable, so this definition tells you exactly what $\mu^{*}$-measurable means. The collection of these sets is a $\sigma$-algebra, and $\mu^{*}$ restricted to that collection is a measure on it. Commented Dec 17, 2015 at 15:41
• Should that not be $\mu^*$ restricted to that collection is an outer measure on it? Commented Dec 17, 2015 at 15:50
• No. $\mu^*$ is an outer measure and restricted to the collection of $\mu^*$-measurable set (wich is a $\sigma$-algebra) it is a measure. Commented Dec 17, 2015 at 17:35

People would just say a set is measurable if Caratheodory condition is satisfied. I feel you always confuse terms. I guess you need change a book, like Royden's

As regard the intuition of Caratheodory condition, it bothered me (and still) and I think it can be a quite deep problem:-( Here is the best explanation I can find here

• Dont you mean outer measurable because it is $\mu^*$? Commented Dec 17, 2015 at 15:50
• $\mu^*$ is a quite standard symbol in real analysis which stands for outer measure. People don't say "outer measurable". They just say a set is measurable if this set satisfies the Caratheodory condition.
– Hua
Commented Dec 17, 2015 at 15:52
• But dont they get confused with the proper measure? Commented Dec 17, 2015 at 15:57
• What do you mean by "they"?
– Hua
Commented Dec 17, 2015 at 15:59
• The expresion $\mu*$-measurable is actually used, especially when there is more than one outer measure involved. Also note that the notation $\mu^*$ for outer measure may be confusing, because it suggests the outer measure may have being defined from a measure / pre-measure $\mu$, and, in fact, that may not be true. There are outer outer measures which are not produced from any measure / pre-measure. Commented Dec 17, 2015 at 21:24