A question about the "dominated" part of the DCT Let $f_n:\mathbb{R}\to [0, \infty)$ be a sequence of measurable functions that converges to some function $f$. Suppose that I show that for "$n$ big enough" we have $|f_n(x)|\le g(x)$ for all $x\in \mathbb{R}$. Am I right in saying that I can still apply the DCT to conclude that $\displaystyle\lim\limits_{n\to\infty}\int_{\mathbb{R}}f_nd\lambda=\int_{\mathbb{R}}fd\lambda$? I think that I can do this simply because I am removing a finite number of terms from my sequence $\displaystyle \int_{\mathbb{R}}f_nd\lambda$ and then I am applying DCT. Is this right?

Suppose that $|f_n(x)| \leq g(x)$ for all $n \geq N$ for some $N \in \mathbb{N}$. Let $(h_n)$ be a sequence defined as $h_k = f_{N+k}$.
Then in particular $h_k(x) \leq g(x)$ for all $k \in \mathbb{N}$. Moreover you clearly have that $h_k \to f$ when $k \to \infty$.
So you can apply the $DCT$ to $(h_k)$ and you obtain that  $$\lim_{k \to \infty}\int_{\mathbb{R}}h_k = \int_{\mathbb{R}}f.$$
Now if you go back to the definition of the limit : let $\epsilon >0$, the above limit implies that there exists $K \in \mathbb{N}$ such that
$$|\int_{\mathbb{R}}h_k - \int_{\mathbb{R}}f|<\epsilon$$
for all $k \geq K$. This implies that
$$|\int_{\mathbb{R}}f_{n} - \int_{\mathbb{R}}f|<\epsilon$$
for all $n \geq K+N$ by definition of $h_k$.
So by definition of the limit, this gives you that
$$\lim_{n \to \infty}\int_{\mathbb{R}}f_n = \int_{\mathbb{R}}f.$$
