1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

F and g are equal almost everywhere iff int(f)dm = int(g)dm for every measurable set.

  1. Apr 1, 2012 #1
    I feel like I got the first part, but the converse is a little tricky. I'm not sure if I am allowed to conclude at the end that f(x) must equal g(x) almost everywhere.

    measurable.png
     
  2. jcsd
  3. Apr 1, 2012 #2

    jgens

    User Avatar
    Gold Member

    Re: f and g are equal almost everywhere iff int(f)dm = int(g)dm for every measurable

    Your argument for the converse does not work. Notice that since [itex]f,g[/itex] are measurable it follows that the sets [itex]E_1 = \{x \in X:f(x) < g(x)\}[/itex] and [itex]E_2 = \{x \in X:g(x) < f(x)\}[/itex] are measurable. Now use the fact that [itex]\int_{E_1} f-g = 0[/itex] and [itex]\int_{E_2} f-g = 0[/itex] to prove that [itex]\mu(E_1 \cup E_2) = 0[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: F and g are equal almost everywhere iff int(f)dm = int(g)dm for every measurable set.
Loading...