Why are some convergent Lebesgue integrals 'undefined'?

Question

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.

Answer 1

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.

Answer 2

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.