Integration

• Dec 12th 2009, 11:53 AM
slevvio
Integration
Using the definition of integration $\displaystyle \int_a^b f(t)dt = sup\{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $\displaystyle g(x) \le f(x) \forall x \in [a,b] \}$ (f continuous on [a,b] )

show that for $\displaystyle \lambda \ge 0, \int_a^b \lambda f(t)dt = \lambda \int_a^b f(t)dt$.

I would really appreciate any help with this. It's easy to show that if lambda is 0 you get the result but I am having trouble proving the rest. I just get tangled up in algebra, though I think I am trying to prove sup A = $\displaystyle \lambda$ sup B, where A = $\displaystyle \{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $\displaystyle g(x) \le \lambda f(x) \forall x \in [a,b] \}$ and B = $\displaystyle \{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $\displaystyle g(x) \le f(x) \forall x \in [a,b] \}$ Thank you very much.
• Dec 13th 2009, 05:44 AM
HallsofIvy
$\displaystyle sup\{\lambda F(x)\}= \lambda sup\{ F(x)\}$

(and, of course, $\displaystyle \int \lambda F(x)dx= \lambda \int F(x)dx$)
• Dec 13th 2009, 07:15 AM
Shanks
provide $\displaystyle \lambda \geq 0$:
for any step function g which supported on [a,b], $\displaystyle g(x) \le f(x) \forall x \in [a,b]$, $\displaystyle \lambda g(x)$ is a step function supported on [a,b], and $\displaystyle \lambda g(x) \le \lambda f(x) \forall x \in [a,b]$.
taking the superem over all step function g, we get that the LHS is no less than the RHS.
and similarly the RHS is no less than the LHS. QED
• Dec 13th 2009, 02:21 PM
slevvio
thank you, worked it out :)