Math, asked by pantiesnlp1233, 5 hours ago

prove that : lambda (A+B)= Lambda A +Lambda B​

Answers

Answered by harshpatel8122006
0

Answer:idk sorry bro

Step-by-step explanation:

Answered by Anonymous
3

Answer:

Prove that for the Lebesgue-measure λ, the inequality

λ(A+B)≥λ(A)+λ(B)

holds.

So, this task is divided in some smaller tasks, and I'm supposed to begin with the following one:

Show that the inequality holds for half-open intervals

A=[a1,b1) and B=[a2,b2)

with a1,a2,b1,b2∈R, b1>a1 and b2>a2.

I tried to do a case analysis, taking a look at A⊂B, A∩B=∅ and A∩B≠∅. Yet, I only came to the conclusion that, no matter what case I try to prove, that there is always only the equality between A and B.

So, we know that A+B=[a1,b1)+[a2,b2)=[a1+a2,b1+b2). This leads to

λ(A+B)=λ([a1+a2,b1+b2))=(b1+b2)−(a2+a1)=b1+b2−a2−a1=b1−a1+b2−a2=λ([a1,b1))+λ([a2,b2))=λ(A)+λ(B).

Of course this doesn't contradict the statement above, but I guess there is more to show here. We are allowed to use the usual properties of a measure.

Edit:

So, basically, I couldn't find an example where the equality didn't hold. I guess this is exactly sub-task wats me to find? We later work with A and B being compact, so I might get a different result, but I'm not sure.

EditEdit:

It turned out that my solution is perfectly fine, I simply refused to believe that the equation would hold for every interval - but it does.

꧁XxMissMasoomxX01 ꧂

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