Can Integrals Take Different Values Without Being Equal?

  • Context: Graduate 
  • Thread starter Thread starter rs123
  • Start date Start date
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
Sign up Log in
1 reply · 2K views
rs123
Messages
4
Reaction score
0
Hi all,

Apologies if this is stupid question, but I have the following situation. Given two measures [itex]u(x)[/itex] and [itex]v(x)[/itex] if [itex]u(x)[/itex] is absolutely continuous to [itex]v(x)[/itex] ( [itex]u<<v[/itex]) I have a result such that

[itex]\int_A f(x)dv(x)[/itex] always takes the value [itex]\int_B g(x)du(x)[/itex]

But strictly [itex]\int_B g(x)du(x)\neq\int_A f(x)dv(x)[/itex]

If that makes sense... Basically I want to express this idea of 'taking the value of without being equal' but don't know how to express it mathematically (in words and symbols).

In case it still isn't clear, performing the first integral will necessarily return the value of the second, but performing the second will never give the value of the first. It appears to be a non symmetric equality.. Or is this nonsense?!

I then want to contrast this to when the measures are equivalent ([itex]u(x)\sim v(x)[/itex]) when the two integrals are equal y'see.

Many thanks!

R
 
Physics news on Phys.org
rs123 said:
Or is this nonsense?!
I'm afraid yes. Equality of values is always symmetric.

If the integrals take a value, then they are equal. E.g. ##\int_0^1 \frac{x}{2}\,dx = \int_0^1 x^3\,dx## even though the functions are different. But as functions, e.g. if the areas of integration ##A,B## are variable, then two functions can be identical almost everywhere: ##F(x) =_{a.e.} G(x)## or ##F(x)=G(x)\;(a.e.)\,.##