Reading over Proposition $3$ of Chapter $4$ several times, I still can't understand a couple of lines in the last paragraph of the proof.
3. Proposition: Let $f$ be defined and bounded on a measurable set $E$ with $mE$ finite. In order that
$$\underset{f\leq \psi}{\text{inf}} \int_E \psi(x) dx = \underset{f\geq \varphi}{\text{sup}} \int_E \varphi(x) dx$$
for all simple functions $\varphi$ and $\psi$, it is necessary and sufficient that $f$ be measurable.
I'm having trouble with the necessary part of the proof:
Suppose now that
$$\underset{f\leq \psi}{\text{inf}} \int_E \psi(x) dx = \underset{f\geq \varphi}{\text{sup}} \int_E \varphi(x) dx$$
Then, given $n$, there are simple functions $\varphi_n$ and $\psi_n$ such that $\varphi_n(x) \leq f(x) \leq \psi_n(x)$ and
$$\int_E \psi_n(x) dx – \int_E \varphi_n(x) dx<\frac{1}{n}.$$
Then the functions $$\psi^* = \text{inf}\: \psi_n$$
$and$ $$\varphi^* = \text{sup}\: \varphi_n$$ are measurable by Theorem 3.20, and
$$\varphi^*(x) \leq f(x) \leq \psi^*(x)$$
Now the set $\Delta= \{x:\varphi^*(x) < \psi^*(x) \}$ is the union of the sets $\Delta_v = \left\{x:\varphi^*(x)<\psi^*(x)-\frac{1}{v}\right\}$
But each $\Delta_v$ is contained in the set $\{x: \varphi_n(x)<\psi_n(x)-1/v\}$, and this latter set has measure less than $v/n$.
I get the build up, but I can't figure out why? Why would the set have measure less than $v/n$? I’ve stared at the page and thought about the problem for a lot of time, but I don’t even know how to start.
Any help is appreciated!
Best Answer
Because $$\int_E\big[ \psi_n(x) - \varphi_n(x)\big] dx<\frac{1}{n}.$$ Which implies $$\int_{\Delta_v}\big[ \psi_n(x) - \varphi_n(x)\big] dx<\frac{1}{n}.$$ And since the integrand is bounded below by $\frac{1}{v}$, the integral is bounded below by $\frac{m(\Delta_v)}{v}$