Thread: 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.

$\displaystyle sup\{\lambda F(x)\}= \lambda sup\{ F(x)\}$

(and, of course, $\displaystyle \int \lambda F(x)dx= \lambda \int F(x)dx$)

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

thank you, worked it out