The lebesgue integral equals the Riemann integral

Question

Suppose that $f$ is integrable on $[a, \infty)$ for some $a \in \mathbb{R}$. Let $\lambda$ denote the restriction of the Lebesgue measure to $[a, \infty)$. Suppose furthermore that $f$ is Riemann integrable over every interval $[a, b]$ with $b > a$. Prove that $$ \int_{[a, \infty)} f ~ d\lambda = \int_a^{\infty} f(x) ~ dx $$ My attempt: we know that there is an increasing sequence $(u_n)_{n \in \mathbb{N}}$ of step functions that converges uniformly to $f$ (from below) and $$ \int_{[a, \infty)} f ~ d\lambda = \sup_{n \in \mathbb{N}} \int_{[a, \infty)} u_n ~ d \lambda $$ However, since $u_n$ is a step function on $[a, \infty)$, it assumes only finite different values on $[a, \infty)$, say the values $r_{n, 1}, \ldots, r_{n, k_n}$. Writing $A_{n, i} = u^{-1}(r_{n, i})$, the disjointness of the $A_{n, i}$ yields $$ \int_{[a, \infty)} f ~ d\lambda = \sup_{n \in \mathbb{N}} \sum_{i = 1}^{k_n} \int_{A_{n, i}} u_n ~ d\lambda = \sup_{n \in \mathbb{N}} \sum_{i = 1}^{k_n} r_{n, i} \cdot \mu(A_{n, i}) $$ But now, I am stuck. What I see is that the right hand side is basically just the interval $[a, \infty)$ split into a finite ($k_n$) amount of pieces and then some lower sum of the integral is computed. Intuitively, this lower sum converges to the integral, but I am unable to prove this.

I have also read something about the Lebesgue and Riemann integral of step functions are identical. However, this would only give me $$ \int_{[a, \infty)} f ~ d\lambda = \sup_{n \in \mathbb{N}} \int_a^{\infty} u_n(x) ~ dx $$ but then again, I am unable to finish the proof.

Any help is greatly appreciated, and solutions involving different methods are welcome as well.

Answer 1

For $n \in \mathbb{Z}_{\geq a}$, define $$ f_n = f \cdot 1_{[a, n]} $$ Then $f_n$ is integrable since $$ \int_{[a, \infty)} |f_n| ~d\lambda = \int_{[a, \infty)} |f \cdot 1_{[a, n]}| ~d\lambda = \int_{[a, n]} |f| ~d\lambda \leq \int_{[a, \infty)} |f| ~d\lambda < \infty $$ But since $f_n$ is integrable and converges to $f$, another integrable function, we have $$ \int_{[a, \infty)} f ~ d\lambda = \lim_{n \to \infty} \int_{[a, \infty)} f_n ~ d\lambda = \lim_{n \to \infty} \int_{[a, \infty)} f \cdot 1_{[a, n]} ~ d\lambda = \lim_{n \to \infty} \int_{[a, n]} f ~ d\lambda = \lim_{n \to \infty} \int_a^n f(x) ~ dx = \int_a^{\infty} f(x) ~ dx $$ where the fourth equality is due to the fact that $f$ is Borel measurable and Riemann integrable over $[a, n]$, which is a compact interval.