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Lebesgue's monotone convergence theorem as stated in Rudin's Principles of Mathematical Analysis (page 318) is the following:

Theorem $\quad$ Suppose $E \in \mathfrak M, \{f_n\}$ is a sequence of measurable functions such that $0\le f_1(x)\le f_2(x)\le ...$ for $x\in E,$ and $f_n\to f$ as $n\to \infty$ on $E$. Then $\int_E f_nd\mu \to \int_E fd\mu.$

I have three simple questions about the applicability of this theorem:

(1) The theorem holds as long as $f_1$ is bounded from below on $E$, right? I.e. $f_1(x)\ge a$ on $E$ for some $a \in \mathbb R$. (In other words, the increasing sequence of functions needs not be non-negative, but just needs to be bounded from below. Simple justification is given at the bottom.)

(2) $f_n$ and $f$ need not be bounded from above, right? (I.e. They may take value of $\infty$ on $E$.) They do not even need to be integrable, do they? (I.e. the integral may be $\infty$.)

(3) Similar statements can be made for a sequence of monotonically decreasing sequence of measurable functions, right? I.e. The theorem holds if $a \ge f_1(x) \ge f_2(x) \ge ...$ for $x \in E, a \in \mathbb R$, and $f_n \to f$ on $E$. $f_n$ and $f$ may take value of $-\infty$ on $E.$

I would appreciate confirmation, refutation and comments/caveats. Thanks a lot!

Justification for (1): Let $g_n(x)=f_n(x)-a$ and $g(x)=f(x)-a$. Then $\int_E g_nd\mu\to \int_E gd\mu,$ as $n\to \infty,$ by the original theorem. But $\int_E g_nd\mu = \int_E (f_n-a)d\mu=\int_E f_nd\mu -a\mu(E)$, and similarly, $\int_E gd\mu =\int_E fd\mu -a\mu(E)$. Hence $\int_E f_nd\mu\to \int_E fd\mu$.

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For (1), no, this is not true unless $\mu(E)<\infty$. In general, it is sufficient for $f_1$ to be bounded from below by an integrable function, by more or less exactly the argument you have given.

For (2), yes, that is correct. This theorem holds also in case of non-integrable limit $f$.

For (3), that is true, with caveats as in case of (1) -- this is easy to see, as if $f_n$ is decreasing, $-f_n$ is increasing, so the results are equivalent by linearity of the integral.

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  • $\begingroup$ Thanks a lot for the answer, and the caveats in particular! Yeah, I missed $\mu(E)< \infty.$ I appreciate your pointing this out! $\endgroup$ – syeh_106 May 15 '17 at 4:02
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    $\begingroup$ The main point is that the integrand may very well become singular, or even just infinite someplace, but then the integral is still well defined: it is just positive infinity, because the negative part is bounded. $\endgroup$ – tomasz May 15 '17 at 5:39

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