Lebesgue integrals help us integrate over unbounded support and not well behaved integrals. we will define them in terms of Riemann integrals. Hubbard & Hubbard has a unique definition of the Lebesgue integral because it does not introduce the notion of measure.
Definition 4.11.1 (Uniform convergence). A sequence $k\mapsto f_k$ of functions $f_k: \R^n \to \R$ converges uniformly to a function $f$ if for every $\epsilon > 0$, there exists $K$ such that when $k\geq K$, then for all $x\in \R^n$, $| f_k(x) - f(x)| < \epsilon$.
Uniform convergence is not a common phenomenon. Polynomial sequences do converge uniformly on a bounded set.
Theorem 4.11.2 (Convergence for Riemann integrals). Let $k \mapsto f_k$ be a sequence of integrable functions $\R^n \to \R$, all with support in a fixed ball $B \subset \R^n$, and converging uniformly to a function $f$. Then $f$ is integrable and
$$ \lim_{k\to \infty} \int_{\R^n} f_k(x) |d^nx| = \int_{\R^n}f(x)|d^nx|. $$
Proof. Choose $\epsilon > 0$ and $K$ so large that $\sup_{x \in \R^n} |f(x) - f_k(x) | < \epsilon$ when $k > K$. Then when $k > K$, we have, for any $N$,
$$ L_N(f) > L_N(f_k) - \epsilon \text{vol}_n(B), \quad U_N(f) < U_N(f_k) + \epsilon \text{vol}_n(B). $$
Now choose $N$ so large that $U_N(f_k) - L_N(f_k) < \epsilon$. We get that
$$ U_N(f) - L_N(f) < U_N(f_k) - L_N(f_k) + 2\epsilon \text{vol}_N(B), $$
yielding $U(f) - L(f) \leq \epsilon (1 + 2\text{vol}_n(B))$.
The following theorem is key to understanding the Lebesgue integral. However, as stated, the hypothesis is rarely satisfied.
Theorem 4.11.4 (Dominated convergence for Riemann integrals). Let $f_k: \R^n \to \R$ be a sequence of R-integrable functions. Suppose there exists $R$ such that all $f_k$ have their support in $B_R(0)$ and satisfy $|f_k| \leq R$. Let $f: \R^n \to \R$ be an R-integrable function such that $f(x) = \lim_{k \to \infty} f_k(x)$ except on a set of measure 0. Then
$$ \lim_{k \to \infty} \int_{\R^n}f_k(x) |d^nx| = \int_{\R^n}f(x) |d^nx|. $$