I know that, if $f$ is Lebesgue integrable, then so is $|f|$. That implies that $g(x)=\sin x/x$ is not Lebesgue integrable in $[0,+\infty[$. However, $g(x)$ is surely Lebesgue integrable in $[0,a]$ for $a>0$. That means that for the Lebesgue integral I cannot have $$\int_{[0,+\infty[}f(x)\:\mathrm{d}x=\lim_{a\to+\infty}\int_{[0,a]}f(x)\:\mathrm{d}x.$$ This raises the question: how do I calculate an unbounded Lebesgue integral? For example $f(x)=e^{-x}$. For the Riemann integral I would approach this calculation in the following way:
$$\int_0^\infty e^{-x}\:\mathrm{d}x=\left. -e^{-x}\right|_0^\infty = \lim_{x\to 0}e^{-x}-\lim_{x\to+\infty}e^{-x}=1.$$
However I feel that I cannot do this in the Lebesgue integral. How could I do it?
The familiar approach still works if you have an absolutely convergent integral: if $\int_0^{\infty} |f| \, dx < \infty$, then the dominated convergence theorem implies that for any increasing sequence $a_n \to \infty$,
$$\lim_{n \to \infty} \int_0^{a_n} f \, dx = \int_0^{\infty} f \, dx.$$
The Lebesgue integral runs into problems when both the positive and negative parts of a function have infinite integrals, so the integral $\int_0^{\infty} \sin x / x \, dx$ does not exist in the Lebesgue sense - only as an improper Riemann integral. Arranging the cancellation carefully is important here.