if f > g on T, then so are the integrals?

this has me confused right now. anyone have any tips on how to prove in general that

if f(x) > g(x) \forall x \in T, then

\int_T f\, dx > \int_T g\, dx

it seems obvious that it's true when you plot it and look at it and there's where my trouble is. how can i get past it being "obvious" and prove it?

The integral calculates the surface under the curve. As f(x) > g(x) \forall x \in T, the surface under g(x) over the interval T must be smaller than the surface under f(x) over the interval T.

Suppose f and g are integrable and f > g for all x in T.

Then f = g + h where h > 0
h = f-g
Integral of h over T is INT F - INT G using linearity of integrals.
Thus, INT H = INT F - INT G
INT F = INT G + INT H

Since H > 0 for all x, any lower sum be positive, and this is less than the value of the integral. Thus INT H > 0 and INT F > INT G.

>>3
Since H > 0 for all x, any lower sum be positive,

Hmmm, orly?  Not 100% sure about that.  Remember H doesn't have to be continuous.

deet beet
futnut

root
>toot

fuck how do I multi-qoute?

>>4
Doh. You're right.

Now I'm thinking that it's not actually true but I'm having trouble producing a counterexample.

What we need to find is a function that's strictly positive that is Riemann-integrable with integral 0. Basically, it'd be something similar to this.

If for rational x = p/q ( in simplest form) you let f( x)  = 1/q, and f( x) = 0 for x irrational, then it can be shown that that f is integrable with value 0. Now what needs to be done is change the irrational values of f so that they are positive, but have a similar pattern to how we defined f at the rationals. Something like that would be a counterexample, though I can't think of a simple way to represent it explicitly. Maybe one could define it based on the decimal expansions.

I'm now pretty sure that what the OP is trying to show is incorrect.

>>8
change the values of f at irrational values of x so that they are positive*

>>7 >>8 I don't think that a counterexample exists.

a function f is Riemann integrable iff f is continous almost everywhere (continuous in R but a set of measure 0). So f is continuous and positive in a set of positive measure, implying that its integral is positive. There is a lot of details inside that i didn't wrote, as im too lazy to write them or translate them, but i'm sure you'll be able to find them.

Ah, so I guess the proof would be beyond me, not having a knowledge of measure and such.

Counterexample: Let T = ∅.  It is obvious that f(x) > g(x) ∀ x ∈ T, f, g.  However, the integrals are zero.

However, if T ≠ ∅, try proving that ∫(f(x) - g(x))dx > 0.

>>8

Functions of real numbers are not Riemann integrable if they have uncountably many discontinuities, so you have to be careful what you set the irrational numbers to.  You can't play the same trick with those that you did with the rationals.

FYI, additionally,

f(x) = 1/q when x = p/q (reduced), a otherwise

is not Riemann integrable unless a = 0.  It is Lebesgue integrable, and has the same integral as f(x) = a.

>>12
Or use any set of measure 0 instead of T.

It's still not clear to me that you can't have a positive function on an interval continuous almost everywhere whose greatest lower bound on any subinterval is 0, which would mean its R. integral is 0.  I slept through this stuff in Analysis, though, so it very well could be true.

what is everyone here retarded, h = f-g is strictly positive and integrable (you can assume almost everywhere, it doesn't matter), hence measurable, ofcourse the integral is also strictly positive

if you want, take an approximating simple function 0 < s < h, integral of s is positive because it's positive on every set of nonzero measure, so integral of h is positive.