I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
We adopt the convention that $0\cdot\infty$ and $\infty\cdot 0$ should both be interpreted to be $0.$
Why is this convention good?
I tried to find a place in which this convention is used.
And I found a place in which this convention is used.
3.1 Definition $\mathcal{S}$-partition
Suppose $\mathcal{S}$ is a $\sigma$-algebra on a set $X.$ An $\mathcal{S}$-partition of $X$ is a finite collection $A_1,\dots,A_m$ of disjoint sets in $\mathcal{S}$ such that $A_1\cup\dots\cup A_m=X.$
3.2 Definition lower Lebesgue sum
Suppose $(X,\mathcal{S},\mu)$ is a measure space, $f:X\to [0,\infty]$ is an $\mathcal{S}$-measurable function, and $P$ is an $\mathcal{S}$-partition $A_1,\dots,A_m$ of $X.$ The lower Lebesgue sum $\mathcal{L}(f,P)$ is defined by $$\mathcal{L}(f,P)=\sum_{j=1}^m \mu(A_j)\inf_{A_j} f.$$
3.3 Definition integral of a nonegative function
Suppose $(X,\mathcal{S},\mu)$ is a measure space and $f:X\to [0,\infty]$ is an $\mathcal{S}$-measurable function. The integral of $f$ with respect to $\mu$, denoted $\int f d\mu$, is defined by $$\int f d\mu=\sup\{\mathcal{L}(f,P):P\text{ is an }\mathcal{S}\text{-partition of }X\}.$$
3.4 integral of a characteristic function
Suppose $(X,\mathcal{S},\mu)$ is a measure space and $E\in\mathcal{S}.$ Then $$\int\chi_E d\mu=\mu(E).$$
Proof
If $P$ is the $\mathcal{S}$-partition of $X$ consisting of $E$ and its complement $X\setminus E$, then clearly $\mathcal{L}(\chi_E,P)=\mu(E).$
$\cdots$
$$\mathcal{L}(\chi_E,P)=\mu(E)\inf_{E} \chi_E + \mu(X\setminus E)\inf_{X\setminus E} \chi_E=\mu(E)+\mu(X\setminus E)\cdot 0.$$
If $\mu(X\setminus E)\in\mathbb{R}$, then $\mu(X\setminus E)\cdot 0=0.$
And by our convention, $\mu(X\setminus E)\cdot 0=0$ even if $\mu(X\setminus E)=\infty.$
I think in this case, our convention was good.
But is this convention always good?
And why is this convention good?
Best Answer
I think it's good in the context of measure theory because it turns $\def\R{\mathbb R}\def\ol{\overline}\ol\R_+ = \R_{\geq0} \cup \{\infty\}$ into a semiring, and it has a monotone continuity property: if $x_n, y_n \in \ol\R_+$ are nondecreasing and $x_n \to x$, $y_n \to y$, then $x_ny_n \to xy$. This enables you to prove the monotone convergence theorem without separating into cases. And it makes the definition of the integral cleaner. Of course you could build measure theory without this convention, but it would make everything more cumbersome. It expresses the fact that the value of a function on a zero-measure set doesn't impact its integral, and that if the function is zero on a set then the integral over that set should be zero, no matter how big the set.
However, outside of this context it's probably a bad idea, since $\cdot$ with the convention is not even continuous on $\ol\R_+$. Maybe somebody can find other contexts in which it's a useful convention, but its usefulness in measure theory seems to be tied to the specific demands of integration.