Proving a Function is Riemann Integrable

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Homework Statement



Let f, g : [a, b] \rightarrow R be integrable on [a, b]. Then, prove that h(x) = max{f(x), g(x)} for
x \in [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
 
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Hint: Prove that if f is integrable then so is |f|.

Why does this help?
 
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:
Try writing max(f,g) in terms of absolute values. To get a feel for this, try to find such an expression for max(f,0).
 
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.
 
max(f,0) = (|f| + f)/2
 
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...
 
Try to think about f-g and f+g.
 
  • #10
ok you I am not getting this

[(g+|g|)+ (f+|f|)] /2 = f+g if f and g are the same sign
 
  • #11
I really don't know how to give you any more hints that don't involve giving you the answer! Maybe try thinking about why (|f|+f)/2 gives us max(f,0). Try to sleep on it.
 
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