If $h: E \rightarrow [0,\infty]$, and $\nu(A) = \int_{A}hd\mu$, then if $A \in \mathcal{A} $ such that $\mu(A) =0 $, we have $\nu (A) =0$

Question

Let $(E, \mathcal{A}, \mu)$ be a measured space and $h: E \rightarrow [0,\infty]$. We define $\nu$ on $\mathcal{A}$ such that $\nu(A) = \int_{A}hd\mu$. Show that if $A \in \mathcal{A}$ verifies $\mu(A) = 0$ then $\nu(A) = 0$

My approach was this: as $h$ is a positive measurable function, there exists a sequence of measurable positive functions $(h_n)$, that is increasing, that is a simple function and that converges to $h$.

We have $h_n \leq h$ and $lim \text{ }h_n = h$, thus by the dominated convergence theorem, we have $\lim \int_A h_nd\mu = \int_A hd\mu$.

As $\int_A h_nd\mu = \sum \alpha_i \mu(A_i)$ where $h_n =\alpha_i$ on $A_i$, and $A_i \subset A$, we have $\mu(A_i)=0$, thus $\int_A h_nd\mu =0$. Because of the limit, we have $ \lim \int_A h_nd\mu = 0$ thus $\int_A hd\mu = 0$.

Thus if $\mu(A) = 0$ we have $\nu(A) = 0$

Is my proof correct?

Answer 1

Seems correct to me apart from two points: Your $h$ is not integrable and only non-negativ. The latter thing is not a problem while the first point requires some change in your proof because you cannot apply the dominated convergence theorem:

Since $h$ is measurable and $h\geq 0$ you can find a monotone sequence of non-negative simple functions $h_n$ such that $$h_n(x)\leq h_{n+1}(x),\quad \lim h_n=h \mbox{ pointwise}.$$ Hence you can apply the monotone convergence theorem and obtain $$\int h_n d\mu \rightarrow \int hd\mu.$$ From this point on your proof seems O.K. to me.

Answer 2

Define $h_n = \min\{h, n\}$ then we have $0\leq h_n \uparrow h$. By monotone convergence theorem $\lim_{n\to \infty} \int_A h_n d\mu = \int_A h d\mu$. Since $0\leq h_n \leq n$ then $\int_A h_n d\mu =0$ if $\mu(A) =0$. From this we have $\int_A h d\mu = 0$.