Let $f:[0,1]\mapsto [0,\infty]$ be a bounded, measurable function, prove that $\int_{[0,1]} f\text { }dm=\inf\int_{[0,1]} \phi\text { }dm$

Question

Today, in class we were presented this theorem

Let $f:[0,1]\mapsto [0,\infty]$ be a bounded, measurable function, prove that $\int_{[0,1]} f\text { }dm=\inf\{\int_{[0,1]}\phi\text{ } > dm\text{ } | \text{ } \phi:[0,1]\mapsto [0,\infty] \text{ is simple, measurable and } f\leq\phi \}$

Our teacher said we can examine the proof if we want to:

I've been trying all day, I get the general idea, but I don't understand the underlined part.

To be clear, what I don't get is why $$\int_{[0,1]}\psi dm\leq\int_{[0,1]}\phi dm$$ implies $$\inf_{f\leq\phi}\int_{[0,1]}\phi dm=\inf_{f\leq \psi\leq M}\int_{[0,1]}\phi dm$$

It would be really helpful if someone explained what theorems, definitions, or reasoning are being used in those underlined parts.

Thanks in advance.

Answer 1

You do not need any theorems here. In measure theory equalities are often proven by proving the $\le$ and $\ge$ cases separately. I'll write $\int f$ for $\int_{[0,1]}fdm$ here for short.

Just before you start, make sure you 100% understand these claims:

• to prove $x\le \inf_A y$ means to prove $x\le y$ for all $y\in A$.
• to prove $\inf_A y\le x$, it is enough to find just some $y\in A$ smaller than the $x$.

Now apply both bullets, in both cases $\le$ and $\ge$:

1. For $\le$, to prove $\inf_{f\le\varphi}\int \varphi \le \inf_{f\le\psi\le M}\int \psi$ means to prove (by the first bullet) $$\inf_{f\le\varphi}\int \varphi \le \int \psi$$ for any $\psi$ with $f\le\psi\le M.$ The infimum in the LHS is smaller than $\int\varphi$ for any $\varphi\ge f$, and in particular for $\varphi:=\psi$ because we just assumed $f\le \psi$. Hence by the second bullet we're done.

2. For $\ge$, to prove $\inf_{f\le\varphi}\int \varphi \ge \inf_{f\le\psi\le M}\int \psi$ means to prove (by the first bullet) $$\int\varphi \ge \inf_{f\le\psi\le M}\int \psi$$ for any $\varphi\ge f$. But as discussed in the previous line in your screenshot, given some $\varphi\ge f$, one can always find a $\psi\le\varphi$ with $f\le\psi\le M.$ With the given $\varphi$ and the $\psi$ found, the inequality is established (by the second bullet): $$\int\varphi \ge \int \psi \ge \inf_{f\le\psi\le M}\int \psi$$

Since each inf is both $\ge$ and $\le$ than the other, they must be equal.