aldrinkleys
- 15
- 0
Did Fubini's Theorem fail here?
Did Fubini's Theorem Fail?
- Context: Graduate
- Thread starter aldrinkleys
- Start date
-
- Tags
- Theorem
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 4K views
Did Fubini's Theorem fail here?
- 22,170
- 3,335
Fubini's theorem didn't fail here, because the conditions are not satisfied.
Roughly, Fubini's theorem (in our case) states that if [tex]f:[0,1]^2\rightarrow [-\infty,+\infty][/tex] is a continuous function, such that one of following conditions holds:
1) [tex]f(x,y)\geq 0[/tex] forall x,y in [0,1]
2) [tex]\iint_{[0,1]^2}{|f(x,y)|dxdy}<+\infty[/tex]
then
[tex]\iint_{[0,1]^2}{f(x,y)dxdy}=\int_0^1{\int_0^1{f(x,y)dy}dx}=\int_0^1{\int_0^1{f(x,y)dx}dy}[/tex]
The problem is that neither (1) nor (2) are satisfied in this case.
Roughly, Fubini's theorem (in our case) states that if [tex]f:[0,1]^2\rightarrow [-\infty,+\infty][/tex] is a continuous function, such that one of following conditions holds:
1) [tex]f(x,y)\geq 0[/tex] forall x,y in [0,1]
2) [tex]\iint_{[0,1]^2}{|f(x,y)|dxdy}<+\infty[/tex]
then
[tex]\iint_{[0,1]^2}{f(x,y)dxdy}=\int_0^1{\int_0^1{f(x,y)dy}dx}=\int_0^1{\int_0^1{f(x,y)dx}dy}[/tex]
The problem is that neither (1) nor (2) are satisfied in this case.
aldrinkleys
- 15
- 0
Thank you!