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Let $(X, \mathfrak{A}, \mu)$ be a measure space. Show that the Monotone Convergence Theorem holds if $f$, $f_{1}$, $f_{2}$, ... are real-valued measurable functions and $f_{1}$ is integrable. Further,

(1) $f_{1} \leq f_{2}...$

and

(2) f = $\lim_{n} f_{n}$ hold almost everywhere.

My try: Assume (1) and (2) hold everywhere. From the monotonicity of the integral we have

$\int f_{1} d\mu \leq \int f_{2} d\mu \leq$ .. $\leq \int f d\mu$.

We need to show the reverse inequality.

Let $g^{+}_{n,k}$ and $g^{-}_{n,k}$ be two sequences of positive valued simple functions such that

$\lim_{k} g^{+}_{n,k} - g^{-}_{n,k} = f_{n}$ and $g^{+}_{n,k} - g^{-}_{n,k} \leq f_{n}$ for all $k$.

Define $h_{n}^{+}$ and $h_{n}^{-}$ as $h^{+}_{n} = \max_{k}(g_{k,n}^{+}$) and $h^{-}_{n} = \max_{k}(-g^{-}_{k,n})$, then both $h_{n}^{+}$ and $h^{-}_{n}$ are non-decreasing measurable simple functions and $\lim_{n} h^{+}_{n} + h^{-}_{n } = f^{+} - f^{-} = f$. Also $h^{+}_{n} \leq f^{+}_{n}$ and $h^{-}_{n} \leq -f^{-}_{n}.$ Since $h^{+}_{n}$ and $h^{-}_{n}$ are simple functions, it follows

$\lim_n \int h_{n}^{+} d\mu = \int f^{+} d\mu$ and $ \lim_{n} \int h^{-}_{n} d\mu = -\int f^{-} d\mu$

and $\lim_{n} \int h^{+}_{n} + h^{-}_{n} d\mu = \int f d\mu$.

Since $\int f^{+} d\mu = \lim_{n} h^{+}_{n} d\mu \leq \lim_{n} \int f^{+}_{n}d\mu$ and similary $-\int f^{-}d\mu = \lim_{n} \int h^{-}_{n} d\mu \leq \lim_{n} -\int f^{-}_{n}d\mu$

$\int f d\mu = \lim_{n} \int h^{+}_{n} + h^{-}_{n} d\mu \leq \lim_{n} \int f^{+}_{n} - f^{-}_{n} d\mu = \int f_{n}d\mu$.

Assume now (1) and (2) hold almost everywhere, let $N$ be the set consisting all $x\in X$ for which at least one of (1) and (2) fails.

Then the functions $f\chi_{n^{C}}$ and $f_{n}\chi_{N^{C}}$ fulfills (1) and (2) and thus

$\int f\chi_{N^{C}} d\mu = \lim_{n} \int f_{n}\chi_{N^{C}}d\mu$ and from this we can conclude that

$\int f d\mu = \lim_{n} \int f_{n}d\mu$.

This is very similar to a proof for MCT in the book Measure Theory by Donald Cohn, however, the book is for positive valued functions only.

My question is whether this is correct or if I have missed something.

Secondly, why is the constraint that $f_{1}$ is integrable necessary?

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One thing as I'm reading it through: "We need to show the reverse inequality." There is no reason why the reverse inequality should hold. Why do you want to show this? –  Thomas E. Feb 28 '13 at 12:29
    
I don't really understand what you mean. We have $\lim_{n}\int f_{n} d\mu \leq \int f d\mu$ and we want to show that $\int f d\mu \leq \lim_{n} \int f_{n} d\mu$ so that $\lim_{n} \int f_{n} d\mu = \int f d\mu$ ? –  Erik Feb 28 '13 at 13:10

1 Answer 1

up vote 3 down vote accepted

Some comments while I'm reading your proof and the original from Donald Cohn's (that you cited). These might sound critical, but I'm just trying to raise your awareness in details.

  • The verse "We need to show the reverse inequality" is a bit puzzling. If, e.g. $(X,\mathcal{M},\mu)=([0,1],\text{Leb}([0,1]),m_{1})$, where $m_{1}$ is the Lebesgue measure, and we choose $f_{n}\equiv 1-\frac{1}{n}$ for all $n\in\mathbb{N}$, then $\int f_{n}\,dm_{1}<\int f_{n+1}\,dm_{1}$ for all $n\in\mathbb{N}$. So there is really no reason why the reverse inequality should hold, and I doubt this is what you meant. You want to show that $\lim_{n} \int f_{n}\,d\mu\geq \int f\,d\mu$.

  • The way you defined the functions $h_{n}^{+}$ and $h_{n}^{-}$ does not work, i.e. defining them as maximums over an infinite set. Are you sure these maximums exist, or should you replace it with a supremum? And if you take supremum, what would be the result? Would these functions, as a supremum of simple functions, be simple functions? In any case, what you want to do is to define $h_{n}^{+}=\max\{g^{+}_{1,n},...,g^{+}_{n,n}\}$ instead, and $h_{n}^{-}$ similarly (but instead minimums).

I am going to assume from now on that the functions $h_{n}^{+}$ and $h_{n}^{-}$ are defined as above. Also it is worth to note that you have not defined $f^{+}$ or $f^{-}$ before you start using these notations.

  • Note that the sequence $(h_{n}^{-})$ does not consist of non-negative simple functions, so you can't use Proposition 2.3.3 as it stands in the next step. Something analogous could be argumented though.

  • And here comes the situation where some level of integrability from $f_{n}$ would be required: \begin{equation*} \lim_{n}\int h_{n}^{+}+h_{n}^{-}\,d\mu=\int f\,d\mu. \end{equation*} What if $\int |f_{n}|\,d\mu=\infty$ for all $n$? Would it be possible that one of the sides would yield $\infty-\infty$ in the steps where you proved this equality, which would not be defined?

Here's a suggestion for an alternative approach to this result, using the MCT for non-negative functions as Donald Cohn has proven it. I don't think you have to make the proof from scratch again.

Take $f_{1}\leq f_{2}\leq ...\leq f$ as before and assume that $f_{1}$ is integrable. Then $0\leq f_{2}-f_{1}\leq f_{3}-f_{1}\leq ... \leq f-f_{1}$ is a non-negative sequence, and thus \begin{equation*} \lim_{n}\int f_{n}\,d\mu-\int f_{1}\,d\mu=\lim_{n}\int f_{n}-f_{1}\,d\mu\overset{MCT}{=}\int f-f_{1}\,d\mu=\int f\,d\mu-\int f_{1}\,d\mu. \end{equation*} Since $-\infty<\int f_{1}\,d\mu<\infty$, then this implies $\lim_{n}\int f_{n}\,d\mu=\int f\,d\mu$.

Edit: Note that the integrability of $f_{1}$ is necessary. Let $(X,\mathcal{M},\mu)=(\mathbb{R},\text{Leb}(\mathbb{R}),m_{1})$ and $f_{n}=-\chi_{[n,\infty)}$ for all $n\in\mathbb{N}$. Now $f_{1}\leq f_{2}\leq ...\leq f=0$, but $\lim_{n}\int f_{n}\,dm_{1}=-\infty\neq 0=\int f\,dm_{1}$.

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Thanks a lot, I agree I was a bit sloppy with some details such as the $\max$-definitions etc, and yes 2.3.3 must be modified, I forgot that. –  Erik Feb 28 '13 at 14:27
    
@Erik. Sure. And you would probably also want to note that $g_{n,k}^{+}$ are chosen to approximate $f_{n}^{+}$ and $g_{n,k}^{-}$ to approximate $f_{n}^{-}$. Even though it is implicitly clear from the steps that follow. –  Thomas E. Feb 28 '13 at 14:29
    
So with the correct definition with $h^{+}$ and $h^{-}$ and a modified version of 2.3.3, is my proof valid? –  Erik Feb 28 '13 at 14:30
    
@Erik. There is one more detail which does not seem correct. Note that $(g_{n,k}^{-})$ increases to $f_{n}^{-}$, whence $(-g_{n,k}^{-})$ decreases to $-f_{n}^{-}$. So you want to take minimums when defining $h_{n}^{-}$ instead of maximums. –  Thomas E. Feb 28 '13 at 14:39
    
Yes, you are right, I was too quick with that. –  Erik Feb 28 '13 at 14:47

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