Proving a Function is Riemann Integrable

  • Thread starter Thread starter SNOOTCHIEBOOCHEE
  • Start date Start date
  • Tags Tags
    Function Riemann
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
10 replies · 9K views
SNOOTCHIEBOOCHEE
Messages
141
Reaction score
0

Homework Statement



Let f, g : [a, b] [tex]\rightarrow[/tex] R be integrable on [a, b]. Then, prove that h(x) = max{f(x), g(x)} for
x [tex]\in[/tex] [a, b] is integrable.
1

Homework Equations



Definition of integrability: for each epsilon greater than zero there exists a partition P so that U(f,P)-L(f,P)<epsilon

The Attempt at a Solution




Ok i have absolutley no clue how to do this one. The following graph is how i think the function would look : http://i31.photobucket.com/albums/c373/SNOOTCHIEBOOCHEE/Graph2.jpg

Sorry about the crude drawing, but the very light blue would be h(x)

But i honest to god can't see a way to make U(f,P)-L(f,P)<epsilon a true statement

Thanks in advance
 
Physics news on Phys.org
I know that that is a theorem in the book, but i don't see how its applicable here. i also know that |Integral(f)| < Integral(|f|) comes from that statement.
 
Last edited:
how do you mean in absolute values?

like max(f,g)<max(|f|,|g|)?

edit: guess not

im trying to figure out this for h(x)= max(f,0)

so we know h(x) = f if f is positive and 0 else.

but i can't figure this out for the other thing.
 
I see where this argument is going. Basically you are gunna describe h(x) as a sum of absolute values of functions we already know are integrable, therefore the result is integrable.

Now to figure out max(f,g)...
 
sorry for the tripple post. but i got a little bit farthermax(f,g)= [(g+|g|)+ (f+|f|)] /2 - min(f,g)

but then i got to figure out how to describe the min function...
 
ok you I am not getting this

[(g+|g|)+ (f+|f|)] /2 = f+g if f and g are the same sign