Is improper integrability equal to integrability in this case?

It is given that $f:\mathbb{R} \to [0,\infty)$ is improperly integrable. I take that to mean that the improper (Riemann) integral $\int_{-\infty}^{\infty} f(x) \, dx $ exists, and, by definition, $f$ is Riemann integrable on any finite interval $[a,b]$ and the limits

$$\lim_{a \to -\infty}\int_a^c f(x) \, dx, \quad\lim_{b\to +\infty}\int_c^b f(x) \, dx,$$

exist and are finite for any $c \in \mathbb{R}$, so that we can define (uniquely)

$$\int_{-\infty}^{\infty}f(x)\, dx = \lim_{a \to -\infty}\int_a^c f(x) \, dx + \lim_{b\to +\infty}\int_c^b f(x) \, dx$$

It also follows that we can evaluate the improper integral (where $n$ is a positive integer) as

$$\int_{-\infty}^{\infty}f(x)\, dx=\lim_{n \to \infty} \int_{-n}^n f(x) \, dx$$

Since $f$ is Riemann integrable on $[-n,n]$, the Lebesgue integral exists over that interval and

$$\int_{\mathbb{R}}f \chi_{[-n,n]} = \int_{[-n,n]}f = \int_{-n}^n f(x) \, dx$$

Since $f\chi_{[-n,n]} \to f$ as $n \to \infty$, and $f$ is nonnegative, we can apply the monotone convergence theorem and obtain

$$0 \leqslant\int_{\mathbb{R}}f = \lim_{n \to \infty} \int_{\mathbb{R}}f \chi_{[-n,n]} = \int_{-\infty}^\infty f(x) \, dx< + \infty$$