I am reading an old book and it says that:

(1) the Lebesgue outer measure has the sub-additivity property (for a countable union of sets $S_i$). OK, but when it says this, it does not require that the sets are pairwise disjoint, actually it explicitly says the sets can be "pairwise disjoint or not".

Then it says that:

(2) the Lebesgue inner measure has the super-additivity property (for a countable union of sets $S_i$) but here it requires that the sets $S_i$ are pairwise disjoint.

This sounds asymmetric to me? Is it really so?

Then in some other Lebesgue measure notes (on the web), I read that even for (1) it is required that the sets are pairwise disjoint.

See Theorem 4.2. here:

Measure Theory Notes

Is it required or not? And if it's not required (but required in (2)), why this asymmetry here between (1) and (2)?! What is the true story? I start to think that even the book may be outdated though I want to accept this only as a last resort.


(4.4.3) outer measure
$\overline L (S_1 + S_2 + ... ) \leq \overline L (S_1) + \overline L (S_2) + ... $

(4.4.6) inner measure
$\underline L (S_1 + S_2 + ... ) \geq \underline L (S_1) + \underline L(S_2) + ... $


1 Answer 1


It is correct as stated. If I cover a set with some $E_i$, I might be overcounting points that appear in more than one $E_i$. For sub-additivity that's fine because double counting points increases the measure.

Measure of union $\leq$ measure of union + measure that was double counted.

What about for super-additivity? If I approximate a set of the inside with some $E_i$, then if I don't double count (i.e. they are disjoint) I should have

Measure of what I covered $\leq$ measure of union

But if my $E_i$ overlap then I can't say

Meaure of what I covered + overlap $\leq$ measure of union

because the overlap could be arbitrarily large. E.g. consider approximating $[0,1]$ with from the inside with $E_i = [1/i, 1]$.

  • $\begingroup$ But if they don't overlap (for super-additivity, for inner measure) isn't it just an equality?! So says intuition at least :) $\endgroup$ Commented Mar 9, 2016 at 22:41
  • $\begingroup$ Only if you cover the whole thing! $[0,1/3], [2/3,1]$ certainly approximate $[0,1]$ from the inside, but I haven't covered the whole thing so I only get inequality. $\endgroup$
    – nullUser
    Commented Mar 9, 2016 at 22:42
  • $\begingroup$ Oh I see what you meant now. Yes for Lebesgue inner measure you get equality (when the set you are covering is measurable because Lebesgue measure is a measure) but for an arbitrary inner measure you only get inequality. $\endgroup$
    – nullUser
    Commented Mar 9, 2016 at 22:44
  • $\begingroup$ Hm... the 2nd comment totally confused me. I just thought I had understood the 1st one. $\endgroup$ Commented Mar 9, 2016 at 22:44
  • $\begingroup$ I don't know yet if the set is measurable or not. We're talking only inner/outer measure here, we haven't said yet if the set (i.e. the union of $S_i$) is measurable or not i.e. if the inner and outer measures of it are equal. We're just looking at them as two separate things for now and we're stating some of their properties. $\endgroup$ Commented Mar 9, 2016 at 22:48

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