# Why is the Monotone Convergence Theorem more famous than it's stronger cousin?

I am reading Stein & Shakarchi. On page 62 we have the Monotone Convergence Theorem:

Suppose $\{f_n\}$ is a sequence of non-negative measurable functions with $f_n\nearrow f.$ Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.

However, just before this theorem we have this much more powerful corrolary of Fatou's lemma:

Suppose $f$ is a non-negative measurable function, and $\{f_n\}$ a sequence of non-negative measurable functions with $f_n(x) \le f(x)$ and $f_n(x) \to f(x)$ for a.e. $x$. Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.

To me, this second corollary seems strictly better than the Monotone Convergence Theorem, yet it is the M.C.T. that has a name and is used often. Am I misunderstanding the theorems, or is there a reason why the M.C.T. is more popular?

Does this corollary have a name?

• Are you sure that corollary is true? I think you would still require integrability of $f$. – Calculon Mar 29 '18 at 9:55
• @Calculon I copied it word for word; I will recheck the proof – Ovi Mar 29 '18 at 9:57
• @Calculon $f$ is required to be non-negative and measurable, so the worst that could happen is that $\int f=+\infty$, in which case the inequality holds trivially. – Arnaud D. Mar 29 '18 at 9:58
• This question seems related : math.stackexchange.com/questions/2468412/… – Arnaud D. Mar 29 '18 at 10:08
• @ArnaudD. Nice catch! – Ovi Mar 30 '18 at 14:57

Some people do call this corollary monotone convergence theorem. Moreover, some instructora introduce this corollary as M.C.T., rather than the origianl version of M.C.T., for instance, Professor Mark J. Schervish and Professor Alessandro Rinaldo at CMU (Lecture notes). It's fine because this corollary is stronger and more practical.

However, from the perspective of mathematics, I think it's better to be traditional and use the original version of M.C.T. and just regard that corollary as a useful corollary.

Let's take a look at the Lebesgue's dominant convergence theorem:

Suppose $\{f_n\}$ and $f$ are measurable functions. If there exists a non-negative function $g$ such that $\int g <\infty$ and $|f_n|\leq g$ a.e., for all $n$. Then $f_n\overset{a.e.}{\rightarrow} f$ or $f_n\overset{\mu}{\rightarrow} f$ implies that $$\lim_{n \to \infty} \int f_n = \int f.$$

Now, compare this with the result you give:

Suppose $f$ is a non-negative measurable function, and $\{f_n\}$ a sequence of non-negative measurable functions with $f_n(x) \le f(x)$ and $f_n(x) \to f(x)$ for a.e. $x$. Then $\displaystyle \lim_{n \to \infty} \int f_n = \int f$.

We can find that when $\int f <\infty$, the result is implied by the Lebesgue's D.C.T. When $\int f =\infty$, by Fatou's lemma we have

$$\infty = \int f = \int \liminf_{n\rightarrow \infty} f_n \leq \liminf_{n\rightarrow \infty} \int f_n\Rightarrow \int f_n\rightarrow \infty.$$

In summary, this result seems more similar to D.C.T than M.C.T. This is one reason why it cannot take the place of M.C.T.

The other reason is more fundamental. If we take a look at the proof of the Fatou's lemma, we may find that it relies on the monotonicity of $g_{k} = \inf_{n\geq k}f_n$ and $g_k\nearrow \liminf_{n\rightarrow \infty} f_n$. That is to say, the Fatou's lemma is actually implied by the M.C.T.

In other words, M.C.T. is the most fundamental theorem, then the Fatou's lemma and finally the D.C.T. Therefore from the perspective of mathematician, it's definitely better to use the "naive version" (actually the only correct version) of M.C.T., but not the seemingly stronger result you give, which is implied by D.C.T. and Fatou's lemma.

Nevertheless, it is true that the result you give here is much more convenient to use than either M.C.T. or D.C.T. That's why some instructors prefer to introduce this result as a substitution for M.C.T.

• Your statement of Lebesgue's dominant convergence theorem doesn't make sense. (Typos?). What does $g$ do? – DanielWainfleet Mar 30 '18 at 0:42
• Thanks a lot for the reminder! Yes it's a typo haha. @DanielWainfleet Now already fixed. – Wanshan Mar 30 '18 at 0:46
• So just to make sure I understand: the result I posted is in fact implied by D.C.T and Fatou's Lemma? Because the top voted answer in the math overflow question gives an example where D.C.T. doesn't work but this theorem works; so I am thinking that that example does not work specifically for Lebesgue measure? mathoverflow.net/questions/296312/… – Ovi Apr 25 '18 at 16:06
• Thanks for the reference and sorry for the late reply. I think for his example we can still use DCT and Fatou to prove that claim. If $\int_X f d\mu<\infty$, we use DCT, otherwise we can use Fatou following the line in my answer to prove that the right hand side of his equality tends to infinity. He said that equality can be used for the definition of Lebesgue integral, but I think it does not make sense because any of the integral theorem depends on the definition of integral, and we cannot use a higher level result to define a lower level concept. @Ovi – Wanshan May 13 '18 at 21:54