Let E be a normed <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-struck">R

Wronsonia8g 2022-07-05 Answered
Let E be a normed R -vector space, ( X t ) t 0 be an E-valued càdlàg Lévy process on a filtered probability space ( Ω , A , ( F t ) t 0 , P ), B B ( E ) with 0 B ¯ and
N t ( ω ) := | { s ( 0 , t ] : Δ X s ( ω ) B } | = s [ 0 , t ] Δ X s ( ω ) 1 B ( Δ X s ( ω ) )
for ω Ω and t 0.

How do we see that t N t ( ω ) is càdlàg?
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Answers (1)

Alexzander Bowman
Answered 2022-07-06 Author has 19 answers
In order for N to be cadlag, it should be assumed that X has finite jump activity in B, i.e. ν ( B ) < , where ν is the Lévy measure of X (otherwise, it blows up to infinity immediately). And in such case, this is rather straightforward: with probability 1, the number of jumps in B is locally finite, and N t = n 1 1 [ τ n , ) ( t ), where τ n is the time of nth jump in B; this is clearly cadlag.

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I was trying to solve the next question:
let X be a finite measure space ( μ ( X ) < ) and let f L 1 ( X , μ ), f ( x ) 0 almost everywhere.
Show that for each measurable subset E X:
lim n E | f | 1 n d μ = μ ( E )
My idea for a solution is to use Fatou's lemma:
In one direction:
E lim inf n | f | 1 n d μ lim inf n E | f | 1 n d μ
E lim inf n | f | 1 n = E 1 d μ = μ ( E )
So we get:
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In the other direction, I thought of maybe saying that we know there is an ε > 0
And an N N so for all n > N we get that 1 + ε > | f | 1 n which means 1 + ε | f | 1 n > 0
and use Fatou's lemma again on the expression above to get the lim sup smaller or equal to μ ( E )

Is it valid? Am I missing something?
If I do, what can I do to prove the other direction?
Thank you!