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Am I correct in assuming, when dealing with Lebesgue integrals on the Cartesian space, that we adopt the notation $\int_{a}^{b} f(x)dx$ where the notation $dx$ is used to denote the Lebesgue measure? However, this notation is identical to the notation of a standard Riemann integral. So, when faced with this sort of notation, how would you know when to solve the integral using the standard Riemann procedure or the Lebesgue procedure? It would make much more sense if the convention for the notation for ALL Lebesgue integrals were to be $\int_{a}^{b} f(x)d\mu(x)$ where $dμ(x)$ denotes the Lebesgue measure.

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    $\begingroup$ The integral of Riemann and Lebesgue are equivalent when the Riemann integral exists. $\endgroup$ – Masacroso Dec 28 '16 at 20:34
  • $\begingroup$ But what if the Riemann integral doesn't exist? Then we're using the same notation regardless? $\endgroup$ – tattybojangler Dec 28 '16 at 20:36
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    $\begingroup$ Another, slightly cumbersome notation that I nonetheless like for its analogy with summation is $\int f(x) \mu(dx)$. One often sees $m$ or $\lambda$ used to denote the Lebesgue measure. Also, the Riemann integral has a notion of direction ($\int_a^b = - \int_b^a$), the Lebesgue integral doesn't. $\endgroup$ – copper.hat Dec 28 '16 at 20:40
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    $\begingroup$ In basic calculus we often treat just the integrals of continuous functions at first. The student may then to go on to general Riemann integrals in another class, usually with more rigor. Do we need different notation then? No. The student might then move on to the Lebesgue theory. Do we need different notation then? No. $\endgroup$ – zhw. Dec 28 '16 at 20:43
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    $\begingroup$ Which we could equally well call an improper Lebesgue integral $\endgroup$ – zhw. Dec 28 '16 at 23:22
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The notation $$ \int_a^b f(x)\;\mathrm dx $$ is not ambiguous. When it can be interpreted both as a Riemann integral and a Lebesgue integral, the integrals coincide.

Integrals such as $\int_0^\infty (\sin x)/x\;\mathrm dx$ are not ordinary Riemann integrals, they are improper Riemann integrals. Although the Lebesgue integral of $(\sin x)/x$ does not exist, the improper Lebesgue integral does exist $$ \lim_{r \to \infty} \int_0^r \frac{\sin x}{x} \;\mathrm dx = \frac{\pi}{2} $$ Moreover, the improper Lebesgue integral coincides with the usual Lebesgue integral when the integrand is absolutely integrable: $$ \lim_{r \to \infty} \int_0^r f(x) \;\mathrm dx = \int_0^\infty f(x)\;\mathrm dx $$ when the right side exists. There is again no confusion for improper integration.

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It doesn't matter which procedure you use because when the Riemann integral exists it is equivalent to the Lebesgue integral. If the Riemann integral doesn't exist it is standard to use the $d\mu$ notation however I have seen the $dx$ notation used so specify the measure although it is generally made clear in the context of the problem that this is the Lebesgue measure.

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