Proof that for a non-negative, bounded, monotonic function, the Riemann and Lebesgue integrals coincide

In particular if $f$ is as described, I want to show that the inproper integrals

$\int_c^\infty f(x) dx$

are the same for Riemann and Lebesgue integral.

My idea is that, on compact intervals i.e. integrating from $c$ to $d>c$ it is true. See e.g. If a function is Riemann integrable, then it is Lebesgue integrable and 2 integrals are the same?.

Then the improper Riemann integral we take the limit of d to infinity.

But this is the same as taking the limit of a lebesgue integral, that is strictly increasing, so we can use monotone convergence.

Is this a correct argument, and is the statement that the improper integrals are equal true?

yes, it is correct. You also have to discuss why $f$ is Lebesgue measurable, which follows from the fact that the inverse image of every interval is an interval. Similarly $f$ is Riemann integrable on every compact set since it is bounded there and its set of discontinuity points is countable, and so it has Lebesgue measure zero. Thus you can talk about improper Riemann integral. Note that in general improper Riemann and Lebesgue integrals do not coincide. The function $f(x)=\frac{\sin x}{x}$ is Riemann integrable in $(0,\infty)$ but not Lebesgue integrable.