Let's assume that $f\colon\mathbb R\to\mathbb R$ is continuous and hence Lebesgue measurable. Then, the Lebesgue integral $\int_{(0,\infty)}f(x)\,d\lambda(x)$ makes sense (but of course can be equal to $\pm\infty$). Given such $f$, how can we find this Lebesgue integral? Will it always be the case that $$\int_{(0,\infty)} f(x)\,d\lambda(x)=\int_0^\infty f(x)\,dx=\lim_{R\to\infty}\int_0^Rf(x)\,dx,\tag{$*$}$$where the second and third integrals are Riemann integrals that can possibly be $\pm\infty$?
I know that if $f$ is such that the improper Riemann integrals $\int_0^\infty f(x)\,dx$ and $\int_0^\infty|f(x)|\,dx$ converge then $f$ is Lebesgue integrable and $\int_{(0,\infty)}f(x)\,d\lambda(x)=\int_0^\infty f(x)\,dx$ us finite.
However, I am asking something different here. If we don't know anything about the convergence of the improper Riemann integrals, do we still have the equality in ($*$)? I know that it is true for bounded interval, and I have studied this result, but I don't see how it generalises.
For a concrete example, let $f(x)=e^x$. This is continuous and hence Lebesgue measurable. Is the following reasoning valid?
The function is Riemann integrable on $(0,R)$ for all $R>0$. Also, since it is Lebesgue measurable its Lebesgue integral makes sense and we have $$\int_{(0,\infty)}f(x)\,d\lambda(x)=\int_0^\infty f(x)\,dx=e^x\bigg|_0^\infty =\infty .$$
In pretty much every source authors interchange between the Riemman integral and the Lebesgue integral automatically. But then, if $(*)$ holds we have $$\int_{(0,\infty)}\frac{\sin x}{x}\,d\lambda(x)=\int_0^\infty \frac{\sin x}{x}\,dx=\frac{\pi}{2},$$ when I've seen written that the Lebesgue integral of $\sin x/x$ over $(0,\infty)$ does not exist. I know that it is not Lebesgue integrable because the integral of $|\sin x/x|$ is unbounded, but why does it not exist?
I think my confusion has something to do with the phrases integral exists, makes sense, and Lebesgue integrable. I use the first two to say that our function has an integral whose value belongs to $[-\infty,\infty]$ and the latter to say that the absolute value of our function has an integral whose value is a non-negative real.
To summarise, if we accept that a Lebesgue integral may be equal to $\pm \infty$, is ($*$) always true for continuous functions $f$? I am not interested in Lebesgue integrability here, I only care for the value of the Lebesgue integral over $(0,\infty)$. I want to use this fact in calculations, as in the example with $f(x)=e^x$.
As you can see I am pretty confused. Any explanation will be appreciated.
Best Answer
$f$ is Lebesgue integrable if it is measurable and $\int|f|<\infty$. That is why $\sin x/x$ is not Lebesgue integrable on $(0,\infty)$, but $e^{-x}$ is. For a measurable function $f$, Lebesgue integrability is equivalent to the existence of $\int|f|$.