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.