# Comparison of the Bounded Convergence Theorem (BCT), Monotone Convergence Theorem (MCT), and Dominated Convergence Theorem (DCT)

I'm trying to understand the relationship between the following theorems:

Bounded Convergence Theorem (BCT):

For a uniformly bounded sequence $$f_n \to f$$ a.e. on a set $$E$$ of finite measure, we have

$$\lim_{n \to \infty} \int_E f_n = \int_E \lim_{n \to \infty} f_n$$

Monotone Convergence Theorem (MCT):

If $$f_n \ge 0$$ and $$f_n \uparrow f$$ a.e., then

$$\int f_n \to \int f$$

Dominated Convergence Theorem (DCT):

If $$f_n \to f$$ a.e., $$|f_n| \le g$$, $$\int g < \infty$$, then

$$\int \lim f_n = \lim \int f_n$$

I'm trying to understand how these convergence theorems are induced from uniform convergence, Egorov's theorem, and Fatou's lemma.

Another topic I want to explore is if these convergence theorems have analogous relationships within sequence continuities (without the integrals), as well as if these convergence theorems involving integrals are subsets/supersets of each other. I would like to find out which cases are more general, and what types of specific cases make one convergence theorem equivalent to another.

• BCT is a special case of DCT
– daw
Apr 22, 2021 at 14:08
• Incidentally: the MCT doesn't need the M. mathoverflow.net/a/296540/37266 Apr 22, 2021 at 22:28

Once you have the MCT, everything else follows.

First, we can show that Fatou's lemma follows from MCT.

Proof: Suppose $$f_n \geqslant 0$$ and define $$g_m = \inf_{k \geqslant m} f_k$$. It follows that $$g_m \leqslant f_n$$ and $$\int g_m \leqslant \int f_n$$ for all $$n \geqslant m$$. Thus, $$\int g_m \leqslant \liminf_{n \to \infty} \int f_n$$. The sequence $$(g_m)$$ is increasing and by definition $$\lim_{m \to \infty} g_m = \liminf_{n \to \infty} f_n$$. By the MCT, it follows that

$$\int \liminf_{n \to \infty} f_n = \int\lim_{m \to \infty} g_m = \lim_{m \to \infty}\int g_m \leqslant \liminf_{n \to \infty} \int f_n\quad \text{(Fatou's lemma)}$$

Then we can show that DCT follows from Fatou's lemma.

Proof: We can assume WLOG that $$f_n \to f$$. (otherwise redefine appropriately on the measure zero set where $$f_n \not\to f$$). Since $$|f_n| \leqslant g$$, we have $$g+f_n \geqslant 0$$. Using Fatou's lemma, it follows that

$$\int g + \int f = \int(f+g) \leqslant \liminf_{n \to \infty}\int(g + f_n) = \int g + \liminf_{n \to \infty}\int f_n,$$

and, hence,

$$\tag{*} \int f \leqslant \liminf_{n \to \infty}\int f_n$$

Similarly, applying Fatou's lemma to $$g- f_n \geqslant 0$$, we get

$$\tag{**} \limsup_{n \to \infty} \int f_n \leqslant \int f$$

Together (*) and (**) imply that

$$\lim_{n \to \infty} \int f_n = \int f \quad \text{(DCT)}$$

• The proof of MCT is independent of any mentioned concepts -- uniform convergence, Egoroff's theorem, etc. It relies on the basic property that if $A_n$ is an increasing sequence of $\mu-$measurable sets, then $\mu\left(\cup_{n=1}^\infty A_n\right ) = \lim_{n \to \infty} \mu(A_n)$.
– RRL
Apr 22, 2021 at 22:16
• Thank you very much for the answer - in your previous comment, do you mean BCT? BCT requires uniform convergence, but I don't think MCT necessarily requires uniform convergence. Apr 23, 2021 at 1:13
• @qxzsilver: You're welcome. I am only connecting MCT with DCT here. By previous comment, do you mean the one immediately after your question by user daw? That says BCT is just a special case of DCT. It appears you are concerned here with Lebesgue integration.
– RRL
Apr 23, 2021 at 2:49
• Uniform convergence as a justification for switching the limit and the integral has more utility when dealing with Riemann integrals (which are defined for finite intervals). In fact, for improper Riemann integrals uniform convergence of $f_n \to f$ on $[a,\infty)$ does not even guarantee that $\lim_{n \to \infty} \int_a^\infty f_n \to \int_a^\infty f$.
– RRL
Apr 23, 2021 at 2:49