# Alternative proof of Monotone Convergence Theorem

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

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