Integration

**slevvio** · Dec 12th 2009, 11:53 AM

Using the definition of integration $\int_a^b f(t)dt = sup\{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $g(x) \le f(x) \forall x \in [a,b] \}$ (f continuous on [a,b] )

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

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 = $\lambda$ sup B, where A = $\{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $g(x) \le \lambda f(x) \forall x \in [a,b] \}$ and B = $\{\int_{- \infty}^{\infty} g(t)dt$ where g is a step function supported on [a,b], $g(x) \le f(x) \forall x \in [a,b] \}$ Thank you very much.

**HallsofIvy** · Dec 13th 2009, 05:44 AM

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

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

**Shanks** · Dec 13th 2009, 07:15 AM

provide $\lambda \geq 0$ :
for any step function g which supported on [a,b], $g(x) \le f(x) \forall x \in [a,b]$ , $\lambda g(x)$ is a step function supported on [a,b], and $\lambda g(x) \le \lambda f(x) \forall x \in [a,b]$ .
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

**slevvio** · Dec 13th 2009, 02:21 PM

thank you, worked it out

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