# Thread: Integral with another measure

1. ## Integral with another measure

Suppose that f is a non-negative integrable function. Let $\displaystyle \lambda (E) = \int _E f dm$ for all $\displaystyle E \in \mathbb M$, the set of all $\displaystyle \sigma$-algebra.

Prove that $\displaystyle \int g d \lambda = \int fg dm \ \ \ \forall g$, g non-negative integrable.

Proof so far.

Now, from an earlier thread, I understand that $\displaystyle \lambda$ is a measure.

Case 1) Suppose that g is a simple function, then $\displaystyle g= \sum ^n_{i=1} a_i 1_{E_i}$, Ei disjoint.

Then $\displaystyle \int g d \lambda = \int \sum a_i1_{E_i}d \lambda = \sum ^n_{i=1} a_i \lambda (E_i)$

Which gives $\displaystyle = \sum _{i=1}^n a_i \int _{E_i}f dm$

Now, is this integrand equals to $\displaystyle \int gf dm$? I think it is but I just can't seem to work it out right.

Thanks.

2. You should simply write $\displaystyle \int_{E_i} f dm= \int 1_{E_i} f dm$, and use linearity of integral like you did before.