#Measure Proof help
33 messages · Page 1 of 1 (latest)
beautiful handwriting
while i would agree mu(\emptyset) = 0 is pretty obvious, i guess you need to write a little more
the rest of the proof makes no sense to me so i’m not even sure what to say about it
do you have a definition of ‘measure’ you can send?
because i think we should start over from there
okay cool
for the empty set property, maybe you could write something like $$f(\emptyset) = \sum_{i \in \bZ\cap \emptyset} 4i = \sum_{i \in \emptyset} 4i = 0$$
slayla
ahh ok ok I see, I realized i didn’t explain that part properly. for the second part of the proof i genuinely had no idea how to prove it 😭
😭
it's okay
so now we have $A_1, A_2, \ldots$ pairwise disjoint sets and we want to show property 2 in the definition. if $$f\left(\bigcup_{i=1}^\infty A_i\right) = \infty,$$ i agree that $$\sum_{i=1}^\infty f(A_i).$$
slayla
but we should probably explain that
and i also want to note that "the infinite union of A_i = \infty" isn't a good way to write that first equality
if you want to write it in as plain english as possible, i would maybe write "the measure of the infinite union of the A_i's is infinity"
ahhh ok ok
me too…. actually i was lost at first because im sure this isn’t a measure
if it’s 4i
Z seems like a typo
or it should be |4i|, or something
i think if it said |4i| instead of 4i, or positive or nonnegative integers instead of Z, that would make f a measure
right right
yea idk, i can carry forward with one of those corrections and continue explaining if you want lol
like if they wrote |4i| instead of 4i.... for this, we have that \cup A_i includes infinitely many integers. otherwise f(\cup A_i) would be finite
i emailed my professor so hopefully that clears it up but thank u!