# Lebesgue Monotone Conv. Theorem's Proof by W. Rudin

The proof is the following

The proposition 1.25 and Theorem 1.19(d) that refers to are these

Why do we need that constant $0 <c <1$ in the proof of Leb. Mon. Conv. Thm? For me, the proof works fine if we take just any simple function $s$ and define the sets $E_n$ with $s$ instead of $cs$.

Thank you! :)

• Do you ask why we take $c<1$ instead of $c=1$? If so imagine that $f=s$ and $f_n$ is strictly small than $s$ then the inequality never holds. – Yanko Sep 14 at 13:10
• Yes! Got it! Thanks – LadyGugu Sep 14 at 13:13
• Oh I already wrote an answer, oh well.. – Yanko Sep 14 at 13:18

If we won't take $c<1$ we couldn't reach the same result. The point is that $f_n$ increasingly convergence to $f$ and so it's integral becomes larger than the integral of any function that is stricly less than f.
If for example $f=1$ (a constant function, say $X=[0,1]$) and $f_n=1-\frac{1}{n}$ then clearly $f_n$ increasing to $f$. But $f$ is a simple function, hence we can take $s=f=1$. In that case for $c=1$ we have that $E_n = \emptyset$ because $s>f_n$ for every given $n$. Therefore the property that $X=\bigcup E_n$ is not satisfied.
Well in the proof, since $0\leq f_n\leq f$ we have trivialy that $$\lim_{n\to \infty} \int f_nd\mu\leq \int fd\mu.$$
given a simple function $\varphi$, $0\leq \varphi\leq f$ all we want, is to get $f_n$ between $\varphi$ and $f$, i.e, $0\leq\varphi\le f_n\leq f$. Maybe this could be hard to get but if we pull $\varphi$ a little bit back :$"\alpha \varphi,~~0< \alpha<1"$ this could be made easier. Well that is the way I like to think about this.