Why isn't Dominated Convergence Theorem taught in intro analysis In a course based off a book like Rudin's Principles of Mathematical Analysis that does non-measure theoretic analysis why isn't dominated convergence taught? It would be useful since continuous functions are Lebesgue measurable. Is there not a way to prove a non-measure theoretic version of this statement? 
 A: I would wager there were well known "dominated convergent theorems" around before the Lebesgue revolution. Two examples come to mind - one, the case of infinite series; this DCT is quite simple to prove, no measure theory needed. The second one concerns integration on an interval. We might for example have a sequence of continuous functions $f_n$ on $(0,\infty)$ that are bounded in absolute value by some $g:(0,\infty) \to [0,\infty).$ We further assume that $\int_0^\infty g <\infty,$ and that $f_n\to f$ uniformly on $[a,b]$ for any $0<a<b<\infty.$ In this case we get $\lim \int_0^\infty f_n \to \int_0^\infty f.$ This is also not hard to prove, and again, no measure theory is needed.
I think it would be good to cover these "easy" DCTs in a Rudin level first course in analysis. Then when students see the big DCT in measure theory, they are ready for it.
A: The usual (Lebesgue) dominated convergence theorem follows from Fatou's Lemma, which is a consequence of monotone convergence theorem. The trick would be to establish some form of the monotone convergence theorem. This is actually not quite obvious without measure theory. A simplified version is sometimes called "Arzela's Dominated Convergence Theorem," which is dominated convergence with the assumption that $|f_n(x)|\leq M$ and that $f_n$ converges to a Riemann integrable function. You can find an "elementary proof" here. There are a couple of proofs, but the crux of most is to establish the notion of countable additivity of measures. This is completely taken for granted in measure theory because it's established in the very beginning! To recall why you'd need this, take a look at the proof of monotone convergence theorem. It requires you to take the limit of the measure of a sequence of increasing sets, whose proof relies on countable additivity. 
A: In short: The Dominated Convergence Theorem is often not taught in introductory analysis, because there exists no direct analogue theorem (see counter-example below).
However, there exist some weaker non-measure-theoretic analysis theorems that are close to the Dominated Convergence Theorem, but they have stronger requirements.
The existence of the Dominated Convergence Theorem is one of the main reasons for the strength of the Lebesgue theory. I recommend reading Should we fly in the Lebesgue-designed airplane? (thanks to EricTowers for pointing that paper out). In the article it is stated:

The value of the Lebesgue theory over the Riemann theory is that it is
superior, as a theory of integration. By this it is meant that
there are theorems in the Lebesgue theory that are true and useful,
but that are not true in the Riemann theory. Probably the most
crucial such theorem is the powerful version of the Dominated
Convergence Theorem that one has in the Lebesgue theory.

Counter-Example
Here is a counter-example why an analogue version of the Dominated Convergence Theorem does not exist:
Consider a countable sequence $(x_n)_{n\in\mathbb{N}}$ from $\mathbb{Q} \cap [0,1]$. and the function defined on [0,1] as
$$ f_n(x) = \begin{cases} 1, & x \in \{x_1,\dots,x_n \} \\ 0, & else \end{cases}$$
Each $f_n$ is Riemann integrable since it is almost everywhere continuous
Further you have $$\int_0^1 f_n(x) dx = 0 $$ for all $n \in \mathbb{N}$.
The sequence $(f_n)_{n\in\mathbb{N}}$ converges monotonically towards
$$ f(x) = \begin{cases} 1, & x \in \mathbb{Q} \cap [0,1] \\ 0, & else \end{cases}$$
However, $f$ itself is not Riemann-integrable and the symbol $$\int_0^1 f \, dx $$ is meaningless and you cannot write
$$ \int_0^1 f \, dx = \lim_{n \to \infty} \int_0^1 f_n(x) dx = 0 .$$
However, since $f \leq 1$ it is clearly Lebesgue integrable you can indeed write the above equation for Lebesgue integrals.
Analogue non-measure theorems:
So this is the Dominated Convergence Theorem:

(Dominated Convergence Theorem) Let $(f_n)_{n=1}^\infty$ be a sequence of measurable functions that
converge pointwise to $f$. Further assume that an integrable function
$g$ exists with $|f_n | \leq g$. Then it holds that $$ \int f dx =
 \lim_{n\to \infty} \int f_n dx  $$

In the case of the Lebesgue Integral, you have that from $|f_n| \leq g $ it follows that when $g$ is integrable, then so is $f_n$. Even further, this also implies that $f$ is integrable, simply because of
$$ \int f \, dx = \lim_{n \to \infty} \int f_n \, dx \leq \int g \, dx < \infty .$$
In the non-measure theoretic analysis this is not true. So if $g$ is Riemann integrable, and you have a sequence $f_n$ which converge pointwise to $f$ with $|f_n| \leq g$, then neither $f_n$ nor $f$ need to be Riemann integrable. And more, even if $f_n$ are Riemann integrable, $f$ is not Riemann integrable  in general. Therefore in the non-measure theoretic analysis, one needs to add requirements that enforce that $f_n$ and $f$ are Riemann integrable.
One possible requirement would be to enforce that all the $f_n$ are bounded by the same constant $M$, Riemann integrable, and that the limit $f$ is also Riemann integrable:

(Arzela’s theorem) If $(f_n)_{n=1}^\infty$ is a sequence of Riemann integrable functions that are bounded by the same constant $M>0$,
definied on a compact interval $I$, and pointwise converge towards an Riemann integrable function
$f$, then it holds $$ \int_I f dx = \lim_{n\to
 \infty} \int_I f_n dx  $$
(For proof, see this link stated from Alex or recent work for easier proof)

Another restriction, uniform convergence, would be to require that the $f_n$ are Riemann integrable and converge uniformly to $f$. This also implies that $f$ is Riemann integrable:

(Well known case) If $(f_n)_{n=1}^\infty$ is a sequence of Riemann integrable functions
definied on a compact interval $I$ which uniformly converge with limit
$f$, then $f$ is integrable and it holds $$ \int_I f dx = \lim_{n\to
 \infty} \int_I f_n dx  $$

