Talk:Measure of Empty Set is Zero

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In order to use the short proof originally given (and commented out), one needs to know that a measure is a finitely additive function, and to do that one has to assume that $\mu \left({\varnothing}\right) = 0$.

Unless I'm wrong about this, and there's a way of proving Countably Additive Function also Finitely Additive without recourse to the fact that $f \left({\varnothing}\right) = 0$.

Actually, the way "countable" is defined on ProofWiki, it means "finite or countably infinite," which means countable additivity is built into the definition. However, even if "countable" is defined as "countably infinite", you can just modify the proof to m(0) = m(0 U 0 U 0 ...) = m(0) + m(0) + ... (by countable additivity) >= m(0) + m(0) (since measures are nonnegative) > m(0), from which it follows that m(0) = m(0) + m(0), implying m(0) = 0.

Your short proof does seem to hang together, so I don't know why I put the long version there. I must have had a reason. No worries, it's gone now. Further comment:
We have Countably Additive Function of Null Set which should demonstrate that $\mu \left({\varnothing}\right) = 0$.
Now I look at it, I see the problem with Definition:Countably Additive Function: it demonstrates it for "finite" but then makes an unjustified leap to "countably infinite".
How are you positioned for tightening it up? I think I might be out of my depth - all I know about this area of mathematics is reading around it. --prime mover 07:01, 21 August 2010 (UTC)


I think I understand now.
The short proof as given here banks on the fact that $\mu \left({\varnothing}\right) = 2 \mu \left({\varnothing}\right)\ $ but that does not rule out the possibility that $\mu \left({\varnothing}\right)$ is transfinite, surely?
Note the original stipulation in the definition of measure that there has to exist at least one set whose measure is finite. Then the long proof is needed, because otherwise (if the measure of all sets is not finite), then the measure axioms (1) and (2) still hold, but what we have defined is not what we normally understand as being a measure.
Hence my construction of the longer proof, which I have reinstated as a separate proof.--prime mover 07:20, 21 August 2010 (UTC)

Whoops, you're right! I feel silly. My apologies. Mag487 06:26, 23 August 2010 (UTC)

Proof 1 has now been removed, as it has been shown to be inadequate. --prime mover 03:51, 22 May 2012 (EDT)