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I sometimes read statements such as

The integral $$\int_0^{\infty} dx \, \frac{\sin x}{x} $$ does not exist as a Lebesgue integral, because it is not absolutely convergent.

But according to my understanding, the integral $$\int_0^{R} dx \, \frac{\sin x}{x} $$ exists as a Lebesgue integral for every $R>0$. Why can't we simply define $$ \int_0^{\infty} dx \, \frac{\sin x}{x} = \lim\limits_{R \rightarrow \infty} \int_0^{R} dx \, \frac{\sin x}{x}, $$ and hence give meaning to the former integral as a Lebesgue integral? Isn't this also how one defines improper Riemann integrals, as a limit of proper integrals? Please point out any misunderstandings.

user111187
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  • The difference is that, in the Lebesgue approach, one does not start with sets of finite width, and then build rectangles from them. One instead starts with finite simple functions bounded by the function, and the domains of these simple functions do not need to be bounded in any way. It is unnatural, from the Lebesgue approach, to restrict the domains of these simple functions first, and then take another limit later based on the width of the domain. – Carl Mummert Nov 04 '14 at 18:49

2 Answers2

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As noted above, for a function $f$ to be Lebesgue - integral, the integral of its absolute value $|f|$ must exist.

That is true not only for Lebesgue - integrals, but whenever we are integrating with respect to a measure (in the classical sense); this is a consequence of the definition of integral (with respect to a measure), which requires that $f$ has a convergent integral in both subsets $f>0$ and $f<0$. That is not the case for the Lebesgue-integral of $\displaystyle{\frac {\sin x} x}$.

The analogue in the discrete case (taking, for instance the counting measure in $\mathbb N$) is that the integrable functions correspond sequences $\{a_n\}$ such that the series $$\sum_n |a_n|<\infty.$$ The absolute convergence of the series guarantees that we can take the sum "in any order we like" and obtain the same value (the integral). This is exactly the same for the integral with respect to any measure (Lebesgue in particular): Integrable functions are those where the value of the integral doesn't depend on the "order of integration". Which is not the case for $$ \int_{\mathbb R} \frac{\sin x} x\ dx. $$ Its discrete analogue would be a series like $$ \sum_{n=1}^\infty (-1)^n \frac 1 n=\lim_{N\to\infty} \sum_{n=1}^N (-1)^n \frac 1 n $$ which although converges, doesn't converge absolutely, causing that rearranging the order of the sum gives you different values (all of which could claim to be the value of the sum, or the integral). Hence, in those cases the integral is not defined.

Alex Ortiz
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Federico
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Although $$ \int_0^R dx \frac{\sin x}{x} $$ exists as a Lebesgue integral for all $R>0$, meaning that $$I(R) \equiv \int_0^R dx \left|\frac{\sin x}{x} \right|$$ has a finite value for all $R$, $$ \lim_{R\rightarrow\infty} I(R) $$ diverges (it goes to infinity roughly like $k\ln R$) , so the Lebesgue integral for $$ \int_0^\infty dx \frac{\sin x}{x} $$ does not exist.

Mark Fischler
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    Why would you take the limit of $I(R)$, instead of taking the limit of the integral without the absolute value signs? – user111187 Nov 04 '14 at 17:51
  • Munmert's comment kind of answers this question. From a coarser viewpoint, the property of a Lebesgue integral is that you are adding thin horizontal rectangles. To cope with functions that may have negative values, the convention is that a rectangle below the X axis counts negative. If the total areas above the axis and below the axis both diverge, then the order in which you group and add the rectangles makes a difference, and indeed you can come up with any answer you like by selecting some order. So Lebesgue integrability to inftyy demands integrals to infty above and below both be finite. – Mark Fischler Nov 04 '14 at 19:07