Proof of Monotone convergence theorem.

Question

In this proof of MCT, I don't understand the the use of the number $\alpha$. When i ignore it and read the proof I don't see where it fails.

Thanks for help.

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Proof.

Let $(X,\mathbb{X},\mu)$ be a measure space.

Then since $f_{n} \leq f$ for all $n$ it follows that $\lim _{n \rightarrow \infty} \int f_{n} \mathrm{d} \mu \leq$ $\int f \mathrm{d} \mu .$

To show the inequality in the other direction we fix a simple function $\phi \leq f, 0<\alpha<1$ and let $$ A_{n}=\left\{x: \alpha \phi(x) \leq f_{n}(x)\right\} $$ Note that for each $n$, $A_{n} \in \mathbb{X}$ and $A_{n} \leq A_{n+1} .$ We also have that $X=$ $\cup_{n=1}^{\infty} A_{n} .$ By Lemma 4.5 we can define a measure $\nu$ on $(X, \mathbb{X})$ by $$ \nu(A)=\int_{A} \phi \mathrm{d} \mu(x) $$

for all $A \in \mathbb{X} .$ It follows from Lemma 3.4 that $$ \int \phi \mathrm{d} \mu=\nu(X)=\nu\left(\cup_{n=1}^{\infty} A_{n}\right)=\lim _{n \rightarrow \infty} \nu\left(A_{n}\right) $$ However for all $n \in \mathbb{N}$ $$ \int_{X} f_{n} \mathrm{d} \mu \geq \int_{A_n} f_{n} \mathrm{d} \mu \geq \alpha \int_{A_{n}} \phi \mathrm{d} \mu=\alpha \nu\left(A_{n}\right) $$ and thus $\alpha \int \phi \mathrm{d} \mu \leq \lim _{n \rightarrow \infty} \int f_{n} \mathrm{d} \mu .$

Since this holds for all $0<\alpha<1$ and simple functions $\phi \leq f$ it follows that $$ \int f \mathrm{d} \mu \leq \lim _{n \rightarrow \infty} \int f_{n} \mathrm{d} \mu .$$

Answer 1

$\alpha$ is there to obtain the result $X=\bigcup_{n=1}^{\infty} A_{n}$.

The reason is as follows.

Since $\ \lim_{n \to \infty}f_n(x)=f(x), \ $ we have $$\lim_{n \to \infty} A_n = \{ x:\alpha \phi(x) \le \lim_{n \to \infty}f_n(x) \}$$ $$ \ \ \ \ =\{ x:\alpha \phi(x) \le f(x) \}.$$

But, for all $x\in X$, we have $f(x) \ge \phi (x)$, so since $\alpha <1, \ $ $f(x) > \alpha \phi (x). \ $ Hence

$$\lim_{n \to \infty} A_n = \{ x:\alpha \phi(x) <f(x) \}.$$

Now, because this is true for all $x\in X$, it follows that $\ \lim_{n \to \infty} A_n = X$.

And since $\ A_1 \subseteq A_2 \subseteq A_3 \subseteq... \ $, we have $\ \bigcup_{n=1}^{\infty} A_{n}= \lim_{n \to \infty} A_n = X$.

Answer 2

$X=\cup_n A_n$ is false if you take $\alpha=1$. For example take $f$ to be a simple function and $\phi =f$. If $f_n$ is strictly increasing to $f$ then $A_n$ is empty.

Proof of the fact that $X=\cup_n A_n$: If $\phi (x)=0$ then $x \in \cup_n A_n$ because $f_n(x) \geq 0$. If $\phi (x) >0$ then $f_n(x) \to f(x)$ and $f(x)\geq \phi (x) >\alpha \phi(x)$ so there exisst $n_0$ such that $f_n(x) >\alpha \phi(x)$ for all $n \geq n_0$. It follows that $x \in A_{n_0}$.

Answer 3

If $\alpha\in(0,1)$ is missing in this proof then it fails because there is no guarantee that in that case $X=\bigcup_{n=1}^{\infty}A_n$.

It is quite well possible that for some $x$ we have: $$\forall n [\phi(x)>f_{n}(x)] $$

Answer 4

[Hope this motivates a bit the introduction of ${ \alpha }$ and sets ${ A _n }$]

Consider a measure space ${ (X, \mathbb{X}, \mu) }.$ Let ${ M ^+ }$ denote the set of all measurable maps ${ X \to [0, \infty] }.$

Th: Let ${ (f _n) }$ be a sequence in ${ M ^+ ,}$ with each ${ f _n \leq f _{n+1} }.$
Then ${ \lim \int f _n d\mu }$ ${ = \int (\lim f _n) d\mu }.$
Pf: i) As ${ f _n \leq f _{n+1} },$ functions ${ f _n \uparrow (\sup _n f _n) }$ pointwise. Also the limit ${ \sup _n f _n }$ is measurable.
As ${ \int f _n \leq \int f _{n+1} },$ integrals ${ \int f _n \uparrow \sup _n \int f _n }.$
We want to show ${ \sup _n \int f _n }$ ${ = \int (\sup _n f _n) }.$ As each ${ \int f _n \leq \int (\sup _n f _n) }$ one has ${ \sup _n \int f _n }$ ${ \leq \int (\sup _n f _n) }.$ So it suffices to show ${ \int (\sup _n f _n) }$ ${ \leq \sup _n \int f _n }.$
We should show ${ \sup \int \varphi }$ ${ \leq \sup _n \int f _n },$ where sup on the left is over all simple measurable maps ${ 0 \leq \varphi \leq (\sup _n f _n) }.$ That is, we should prove ${ \int \varphi }$ ${ \leq \sup _n \int f _n }$ for every simple measurable map ${ 0 \leq \varphi \leq (\sup _n f _n) }.$
ii) Let ${ \varphi }$ be a simple measurable map with ${ 0 \leq \varphi \leq (\sup _n f _n) }.$ To show ${ \int \varphi }$ ${ \leq \sup _n \int f _n },$ we could show ${ \alpha \int \varphi }$ ${ \leq \sup _n \int f _n }$ for every ${ \alpha \in (0,1) }.$
So let ${ \alpha \in (0,1) }$ be arbitrary. We have sets ${ A _n }$ ${ := \lbrace x \in X : \alpha \varphi (x) \leq f _n (x) \rbrace }$ : These are measurable, and over them ${ \int _{A _n} (\alpha \varphi) }$ ${ \leq \int _{A _n} f _n \text{ } (\leq \int _X f _n) }.$
We see ${ A _n \subseteq A _{n+1} },$ also ${ X = \cup A _n }.$ (Because say ${ x \in X }.$ If ${ \varphi (x) = 0 }$ then ${ x \in \cap A _n }$ anyways, so suppose ${ \varphi(x) \gt 0 }.$ Now ${ \alpha \varphi (x) }$ ${ \lt \varphi(x) }$ ${ \leq (\sup _n f _n) (x), }$ so ${ \alpha \varphi (x) \lt f _{N} (x) }$ for some index ${ N },$ giving ${ x \in \cup A _n }$).
So ${ \int _{A _n} (\alpha \varphi) \uparrow \int _X (\alpha \varphi) },$ and since each ${ \int _{A _n} (\alpha \varphi) \leq \int _X f _n }$ taking ${ n \to \infty }$ gives ${ \int _X (\alpha \varphi) }$ ${ \leq \sup _ n (\int _X f _n )}$ as needed.