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My current lecture states the theorem of monotone convergence:

Let $(X, \mathcal{E}, \mu)$ a measure space and $\varphi_n: X \rightarrow \mathbb{R}$ an increasing sequence of $\mu$-integrable functions with:

$$\exists M \in \mathbb{R}: \forall n \in \mathbb{N}: \int_X\varphi_n\,d\mu \leq M$$

Then $\varphi := \lim \varphi_n: X \rightarrow \overline{\mathbb{R}}$ is $\mu$-integrable with:

$$\int_X \varphi\,d\mu = \lim_{n \rightarrow \infty} \varphi_n\,d\mu$$

Note: A function $f: X \rightarrow \mathbb{R}$ is called $\mu$-integrable iff $\int_X|f|\,d\mu < \infty$.

Wikipedia states:

Let $(X, \mathcal{E}, \mu)$ be a measure space. For a pointwise non-decreasing sequence of $\mathcal{E}$-measurable non-negative functions $f_k: X \rightarrow [0, \infty]$ consider the pointwise limit $$f := \lim_{k\rightarrow\infty} f_k$$ Then $f$ is $\mathcal{E}$-measurable with:

$$ \lim_{k\rightarrow\infty} \int_X f_k\, d\mu = \int_X f\,d\mu$$

Now I'm a little bit confused my lecture doesn't require the functions of the sequence to be non-negative but introduces the extra upper bound. What is the difference here?

The Wikipedia definition allows it to set the integral of the limit equal to the limit of the integrals. The lecture definition permits this only under the condition, that the sequence is bounded. Is this really required?

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  • $\begingroup$ Can $\int_X \varphi_n d\mu = -\infty$? $\endgroup$ – BigbearZzz Jan 14 at 14:38
  • $\begingroup$ @BigbearZzz No, since it's required that the $\varphi_n$ are $\mu$-integrable (the integral of the absolute value is not $\infty$) $\endgroup$ – user7802048 Jan 14 at 15:07
  • $\begingroup$ So you're using the word $\mu$-integrable in the sense that $\int_X |\varphi| d\mu < \infty$? I just want to make sure because I've seen people use it differently. $\endgroup$ – BigbearZzz Jan 14 at 15:13
  • $\begingroup$ @BigbearZzz Yes exactly. I'll edit my question to include this information. $\endgroup$ – user7802048 Jan 14 at 15:15
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Since $\varphi_1$ is assumed to be $\mu$-integrable, we have $$ \int_X \varphi_1^- d\mu <\infty, $$ where $f^-(x) := \max\{0,-f(x)\}$ denotes the function's negative part. We can consider instead the sequence $\psi_n:= \varphi_n+\varphi_1^-$. Obviously, we have $\psi_1=\varphi_1^+\ge 0$ and since $\varphi_n$ is increasing, $\psi_n\ge0$ for all $n\in\Bbb N$.

Since $\varphi_n \to \varphi$, we have $\psi_n\to \psi = \varphi + \varphi_1^-$. By the usual monotone convergence theorem, we can deduce $$ \int_X \psi\, d\mu = \lim_{n\to\infty} \int_X \psi_n \,d\mu $$ which means that $$ \int_X \varphi\, d\mu + \int_X \varphi_1^-\, d\mu = \lim_{n\to\infty} \int_X \varphi_n + \varphi_1^-\,d\mu $$ can we can subtract $\int_X \varphi_1^- \,d\mu$ from both sides to get $$ \int_X \varphi\, d\mu = \lim_{n\to\infty} \int_X \varphi_n \,d\mu. $$

I don't see why you'd need the bound $\int_X \varphi_n d\mu<M$ at all.

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  • $\begingroup$ Thank you for the detailed answer! $\endgroup$ – user7802048 Jan 14 at 16:03

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