Proof of The Integral of a Multiple by a Constant in [0,∞], the Constant ∞ Included

Real $c$. If $c\neq\infty$, then $c$ is a real number with $c\ge0$, and the claim is Basic Properties of the Integral of Nonnegative Measurable Maps (iv) with $h=cf$.

The case $c=\infty$. For $k\in\mathbb{N}$ let $f_{k}=kf$. By the real case, $f_{k}\in\mathcal{M}^{+}$ and $\int_{X}f_{k}\,d\mu=k\int_{X}f\,d\mu$. Let $x\in X$. Since $k\le k+1$, Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (iv) gives $f_{k}(x)=k\,f(x)\le(k+1)\,f(x)=f_{k+1}(x)$. Moreover $\mathbb{N}\subseteq[0,\infty]$ is not bounded above in $\mathbb{R}$ (Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema (ii)), so $\sup\mathbb{N}=\infty$ by Order and Suprema in the Extended Real Line (v), and with the commutativity of the product (Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii)) and Suprema and Finite Sums in [0,∞] (ii),

$$\sup_{k}f_{k}(x)=\sup\{f(x)\,k:k\in\mathbb{N}\}=f(x)\cdot\sup\mathbb{N}=\infty\cdot f(x)=(\infty f)(x).$$

So the functions $f_{k}$ increase pointwise to $\infty f$, and the monotone convergence theorem, Monotone Convergence Theorem (i) and Monotone Convergence Theorem (ii), gives $\infty f\in\mathcal{M}^{+}$ and

$$\int_{X}\infty f\,d\mu=\sup_{k}\int_{X}f_{k}\,d\mu=\sup_{k}\Bigl(k\int_{X}f\,d\mu\Bigr)=\infty\cdot\int_{X}f\,d\mu ,$$

the last equality by the same computation with $\int_{X}f\,d\mu$ in place of $f(x)$.