A question on relating Riemann-Stieltjes integral and Lebesgue integral

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I am having trouble on understanding an argument in the book "Measure and Integral" (second edition) by Wheeden-Zygmund. The argument is in the proof of Theorem 5.43 (page 100). For convenience let me state the theorem here.

Theorem 5.43. If either $\int_Ef$ or $\int^{+\infty}_{-\infty}\alpha d\omega(\alpha)$ exists and is finite, then the other exists and is finite, and $\int_Ef=-\int^{+\infty}_{-\infty}\alpha d\omega(\alpha)$.

Here $f$ is a measurable function defined on the measurable set $E$ and is finite almost everywhere, and $\omega$ is the distribution function of $f$.

The proof goes as follows. First of all the authors quote Theorem 5.42, stating that $\int_{E_{ab}}f=-\int^b_a\alpha d\omega(\alpha)$. Then let $f\in L^1(E)$. The authors claim that as $a\to-\infty$ and $b\to+\infty$, we have $\int_{E_{ab}}f\to\int_Ef$. The authors say it is true because this holds for both $f^+$ and $f^-$. This is the argument I don't understand; I don't see why this limit holds for both $f^+$ and $f^-$. Actually I don't even see why the limits $\int^{+\infty}_{-\infty}f^+$ and the one for $f^-$ exist.

Thank you.

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The exact answer to this question depends on how the Lesbegue integral is defined in your text. There are many variants, and I don't know which this book uses. Perhaps the most common is that for $f \ge 0, \int_E f := \max \{\int_E s \mid s \le f \text{ is a step function}\}$, and for general $f$, $$\int_E f := \int_E f^+ - \int_E f^-$$

By this definition, if $\int_E f^+$ and $\int_E f^-$ are both $\infty$, then $\int_E f$ would be undefined. If either were infinite while the other was finite, then $\int_E f$ would be infinite. Therefore for $\int_E f$ to be finite, it must be that both $\int_E f^+$ and $\int_E f^-$ are finite.

To show that for $g \ge 0, \int_{E_{ab}} g \to \int_E g$ as $a \to -\infty, b \to \infty$, let $$\chi_{ab}(t) := \begin{cases}1 &t\in[a,b]\\0&t\notin[a.b]\end{cases}$$ be the characteristic function. Note that $g\chi_{ab} \nearrow g$ and $\int_E g\chi_{ab} = \int_{E_{ab}} g$. Now apply the Dominated Convergence theorem.