Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a non negative Lebesgue measurable function. For any set $A\in \mathcal{M}$ ($ \mathcal{M}$ is the $\sigma$-algebra of Lebesgue measurable sets), define $\mu(A):=\int_A f d\lambda$.
Show that $\mu$ is also a measure on $\mathcal{M}$

Question

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a non negative Lebesgue measurable function. For any set $A\in \mathcal{M}$ ($ \mathcal{M}$ is the $\sigma$-algebra of Lebesgue measurable sets), define $\mu(A):=\int_A f d\lambda$. 
Show that $\mu$ is also a measure on $\mathcal{M}$

zzr0ck3r · Answer

@eliassaab this will be the last one ever:) until maybe graduate school:)

anonymous · Answer

It is clear that 
$$
\mu(A) \ge 0\
\mu(\emptyset)=0\

$$
We need to show that $ \mu $ is $\sigma $-additive.

anonymous · Answer

Let $ A_n$ be a sequence of pairwise disjoint measurable sets and let $A=\cup_{n=1}^\infty A_n   $. We need to show that $$
\mu (A) =\sum_{n=1}^\infty \mu (A_n)

$$

anonymous · Answer

Let $ B_k=\cup_{n=1}^k A_n$. It is clear that $B_k$ is increasing, the union of all 
$ B_k$ is equal to A and
$$
\mu(B_k)=\int_{B_k} f d\lambda=\sum_{n=1}^k \int_{A_n} f d\lambda= \sum_{n=1}^k 
\mu(A_n)
$$
By a problem that we did before we have that
$$
\lim_{k\to\infty}\int_{B_k} f d\lambda=\int_A fd\lambda=\mu(A)=\
\lim_{k\to \infty } \sum_{n=1}^k 
\mu(A_n)=\mu(A)\
\sum_{n=1}^\infty 
\mu(A_n)=\mu(A)
$$
Do not worry, you can always ask me problems.