Understanding proof that a integrable function on general measure space is finite almost everywhere

Solution 1:

Fleshing out @Gribouillis comment, a proof is as follows:

Note that $\chi_{X_\infty} \leq f$, where $\leq$ is defined pointwise, and $f|_{X_\infty} = \infty$. So $n \chi_{X_\infty} \leq f$ on $X$, for every $n$. So we have \begin{equation} n \mu(X_\infty) = \int n \chi_{X_\infty} \leq \int f < \infty. \end{equation} Rearranging we have \begin{equation} \mu(X_\infty) \leq \frac{1}{n} \int f \overset{n \to \infty}{\to} 0, \end{equation} and we are done.

Another proof that seems to be suggested by how Royden writes it is to first show that the formula for calculating the integral of a simple function still holds when we allow the function to take on the value $\infty$. Then, we have \begin{equation} \int f = \int_{X\setminus X_\infty} f + \int_{X_\infty} f < \infty \end{equation} Then since $f|_{X_\infty} = \infty$, we would have \begin{equation} \int_{X_\infty} f = \infty \cdot \mu(X_\infty) \end{equation} which is infinite if $\mu(X_\infty)>0.$ hopefully this proof sketch helps get some feel for it.