showing that $f: R^n \mapsto R\cup\{\pm\infty\}$ is Lebesgue integrable

Question

Consider $f: R^n \mapsto R\cup\{\pm\infty\}$ a function. If I am asked to show that $f$ is Lebesgue integrable, is it enough to prove that : $$\int_{R^n}f(x)dx<\infty$$ or do I have to show that $f$ is in $S^{\uparrow}$ which means that $f$ is the limit of functions $(f_k)$ with $f_k\leq f_{k+1}$ for all $k\in N$ and $f_k$ are continuous functions with compact support... I am asking this because the criteria that we were given in class is that $f$ is Lebesgue integrable if and only if the inferior integral of $f$ equal the superior integral of $f$ and both are finite.

Answer 1

Let $(X, \mathbb{X}, \mu)$ be a measure space and $L^{+}(X):= \{f: X \to [0,+\infty], \hspace{0.1cm} \mbox{$f$ is measurable}\}$. If $f \in L^{+}(X)$, then its integral is just: $$\int f d\mu := \sup \{\int\phi d\mu, \hspace{0.1cm} \mbox{$\phi \in L^{+}(X)$ and $\phi$ is a simple function such that $\phi \le f$}\}$$ Note that this does not prevent your integral to be infinite. If, on the other hand, $f: X \to \mathbb{R}\cup\{\pm \infty\}$, then you have to adapt the previous definition. One usually defines $f^{+}(x) := \max\{0, f(x)\}$ and $f^{-}(x) := \max\{0, -f(x)\}$. Note that $f^{+}$ is just the 'positive part' of $f$ while $f^{-}$ is the 'negative part' of $f$ but with positive sign. Thus, both $f^{+}$ and $f^{+}$ are now in $L^{+}(X)$. Then, we define $f$ to be integrable if: $$ \int f^{+}d\mu < +\infty \quad \mbox{and} \quad \int f^{-}d\mu < +\infty $$ These conditions prevent some undefined operations such as $+\infty - \infty$. If both the above conditions are satisfied, we define: $$\int f d\mu := \int f^{+}d\mu - \int f^{-}d\mu $$

Thus, if you want to show that $f$ is integrable it is enough to show that $|f|$ has a finite integral. In fact, note that $|f| \equiv f^{+}+f^{-}$, so that: $$\int |f| d\mu < \infty \Rightarrow \int f^{+}d\mu + \int f^{-}d\mu < +\infty \Rightarrow int f^{\pm}d\mu < +\infty$$ so that $f$ is integrable.

It is worth mentioning that the converse is also true: if $f$ is integrable, then the integral of $|f|$ must be finite.