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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.

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1 Answer 1

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$f$ 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 $\int|f|$.

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  • $\begingroup$ Thanks for your quick answer, Yes I understand this. But that was not my question. My question was if we accept that a Lebesgue integral may be equal to $\pm\infty$, is (∗) always true for continuous functions $f$? Is the value the same? $\sin x/x$ may not be Lebesgue integrable, but it is continuous on $(0,\infty)$ and hence measurable. So its integral "exists". What is its value then? $\endgroup$
    – Zugzwang14
    Commented Dec 1, 2015 at 22:30
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    $\begingroup$ The Lebesgue integral is defined first for functions of constant sign. For a general $f$, inis defined as $\int f=\int f^+-\int f^-$, where $f^+$ and $f^-$ are the positive and negative part of $f$. If $\int f^+=+\infty$ and $\int f^-=-\infty$, then $\int f$ is not defined. $\endgroup$ Commented Dec 1, 2015 at 22:38
  • $\begingroup$ Okay, then if $\int f^+$ and $\int f^-$ are bounded, and $f$ is continuous, does $(*)$ hold? $\endgroup$
    – Zugzwang14
    Commented Dec 1, 2015 at 22:41
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    $\begingroup$ If $\int_0^\infty|f|<\infty$ and $f$ is Riemman integrable on bounded intervals, then the Lebesgue integral coincides with the improper Riemann Integral. $\endgroup$ Commented Dec 1, 2015 at 22:55
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    $\begingroup$ If $f$ is continuous then it is Riemman integrable on bounded domains, so the answer is yes, that's how yo do it. $\endgroup$ Commented Dec 2, 2015 at 10:00

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