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From what I did study, I know that Lebesgue integral is more general than Riemann integral. Then, does Lebesgue integral satisfy all of the Riemann integral properties?

In particular, is the following true? For some given set X,

$\int_{A}f d\mu + \int_{B}f d\mu = \int_{A\cup B}f d\mu$

where $M$ is a $\sigma$-algebra in a set $X$ and $\mu$ is a positive measure on $M$.

BigbearZzz
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J.U.math
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    What if $A=B$? You have forgotten something ... – user251257 Aug 18 '16 at 20:37
  • This would only hold if $A\cap B=\emptyset$. But yes in that case it does. – Cupitor Aug 18 '16 at 20:37
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    To answer your question: any Riemann integrale function on a compact interval is also Lebesgue integrable and their Integrals agree. So obviously the Lebesgue integral has all the properties of the Riemann integral restricted to Riemann integrable functions on compacts. – user251257 Aug 18 '16 at 20:39
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    Lebesugue Integral is genenral. – Seongqjini Aug 18 '16 at 21:42

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As was stated in the comments, if the domain is compact then the Riemann and Lebesgue integral agree. But one thing Riemann has over Lebesgue is that it allows improper integrals. This question from 2013 gives an example: Riemann-integrable (improperly) but not Lebesgue-integrable

Community
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B. Goddard
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    I don't know if this is really true in the sense that you could do the same with Lebesgue-integrals. I.e. you could just look at the limits $\lim_{n \to \infty} \int_{[k,n]} f(x) \mathrm{d}x$. –  Aug 18 '16 at 21:05
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    What stops us from defining improper Lebesgue integrals, in exactly the same we we do for improper Riemann integrals? –  Aug 18 '16 at 21:08
  • This has been discussed to death. See also 2011: http://math.stackexchange.com/q/67198 – B. Goddard Aug 18 '16 at 21:19
  • Yes, there's no dispute that the (proper) Lebesgue integral of, for example, $\sin(x)/x$ does not exist. But nothing stops us from defining an improper Lebesgue integral as @menag indicated. –  Aug 18 '16 at 21:31
  • @Bungo improper Lebesgue integral is very unusual, as it won't agree with the Lebesgue integral and doesn't have it's properties. However, improper absolutely Lebesgue integrable is the same as proper integrable by the convergence theorems. – user251257 Aug 18 '16 at 23:02
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    @user251257 No question, you have to take be careful with any improper integral (Riemann or Lebesgue). My only point is that the ability to define an improper integral does not seem to be a special advantage of Riemann over Lebesgue as claimed in this answer. Certainly we can use either integral to assign a meaningful value to $\int_{-\infty}^{\infty}\sin(x)/x$, with the understanding that, as with any conditionally convergent series, the value depends on the order in which the terms are added, regardless of which integral we use. –  Aug 18 '16 at 23:07
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    Indeed, it is the Lebesgue integral that enjoys the advantage over Riemann with respect to improper integrals, because in the absolutely convergent case, Lebesgue doesn't require an improper integral whereas Riemann does. –  Aug 18 '16 at 23:13