# The lebesgue integral equals the Riemann integral

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.

• Do you know (and can you use) that $\int_{[c,d]} g\,d\lambda = \int_c^d g(x)\,dx$ for $c < d < +\infty$ and all functions $g$ that are both Riemann and Lebesgue integrable over $[c,d]$? Commented Dec 7, 2016 at 14:35
• Yeah but that's kind of what I have to prove, right? Commented Dec 7, 2016 at 14:37
• You want to prove the analogous result for $[a,+\infty)$. I ask whether you can use the fact for compact intervals in the proof. Commented Dec 7, 2016 at 14:38
• Unfortunately, I don't think that that is what I'm supposed to do. Commented Dec 7, 2016 at 14:38
• @DanielFischer I just asked, and it seems that we can use this result. Sorry for the mistake. This immediately gave me what I needed. Commented Dec 7, 2016 at 15:35

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.

• It's not sufficient that $f_n$ and $f$ are integrable and $f_n$ converges (pointwise) to $f$ to conclude $\int f_n\,d\lambda \to \int f\,d\lambda$. But here we have $\lvert f_n(x)\rvert \leqslant \lvert f(x)\rvert$ for all $x$, so the dominated convergence theorem guarantees the result. Commented Dec 7, 2016 at 16:22