# Math Help - Integration II

1. ## Integration II

Let $\lambda \in \mathbb{R}, f:[a +\lambda , b + \lambda ] \rightarrow \mathbb{R}$ continuous.

Is this correct?

$\int_{a+ \lambda}^{b+\lambda} f(s)ds := sup \{ \int_{-\infty}^{\infty} g(s)d(s):$ g supported on $[a +\lambda , b + \lambda ], g(s) \le f(s) \forall s \in [a +\lambda , b + \lambda ] \}$ = supA.

Define $h(s) = g(s+\lambda)$. Then h is supported on [a,b], $h(s) \le f(s + \lambda) \forall s \in [a, b]$.

Hence A = $\{ \int_{-\infty}^{\infty} h(s)d(s):$ h supported on $[a , b ], h(s) \le f(s+\lambda) \forall s \in [a , b ] \}$ = B. [since $\int_{-\infty}^{\infty} h(s)d(s) = \int_{-\infty}^{\infty} g(s)d(s)$ ??? ]

so sup A = sup B $\iff \int_{a+ \lambda}^{b+\lambda} f(s)ds = \int_{a}^{b} f(t+\lambda)dt$. ?

Thanks for any advice or input