Question 10.T.17: If the functions f, g : Ω → ¯R are measurable, then each of ...
If the functions f, g : Ω → \bar{\mathbb{R}} are measurable, then each of the following functions is also measurable:
(i) f + g : Ω_{0} → \bar{\mathbb{R}}, where
Ω_{0} = \left\{x ∈ Ω : (f (x), g(x)) ≠ (∞, −∞), (−∞, ∞)\right\}.
(ii) cf : Ω → \bar{\mathbb{R}}, where c ∈ \mathbb{R}.
(iii) |f | : Ω → [0, ∞].
(iv) f^{n} : Ω → \bar{\mathbb{R}}, where n ∈ \mathbb{N}.
(v) 1/f : Ω → \bar{\mathbb{R}}, where (1/f )(x) = ∞ if f(x) = 0.
(vi) f g : Ω → \bar{\mathbb{R}}.
(i) Note that Ω is replaced here by Ω_{0} in order to avoid the indeterminate forms ∞−∞ and −∞+∞. Since the sets f^{−1}(\left\{∞\right\}), f^{−1}(\left\{−∞\right\}), g^{−1}(\left\{∞\right\}), and g^{−1}(\left\{−∞\right\}) are all measurable, Ω_{0} is itself measurable.
Let α ∈ \mathbb{R} and consider the set
A = \left\{x ∈ Ω_{0} : f (x) + g(x) < α\right\}
= \left\{x ∈ Ω_{0} : f (x) < α − g(x)\right\}.
Since \mathbb{Q} is dense in \mathbb{R}, for every x ∈ Ω there is a q ∈ \mathbb{Q} such that f(x) < q < α − g(x). Consequently
A = \underset{q∈\mathbb{Q}}{\cup} \left\{x ∈ Ω_{0} : f (x) < q\right\} ∩ \left\{x ∈ Ω_{0} : g(x) < α − q\right\}.
But for each q ∈ \mathbb{Q}, the set \left\{x ∈ Ω_{0} : f (x) < q\right\}∩ \left\{x ∈ Ω_{0} : g(x) < α−q\right\} is measurable, hence A ∈\mathcal{M}.
(ii) If c = 0 then the function cf = 0 on Ω is measurable because it is continuous. Suppose, therefore, that c ≠ 0. In this case, for all α ∈ \mathbb{R},
{x ∈ Ω : cf(x) > α} = {x ∈ Ω : f(x) > α/c} if c > 0,
{x ∈ Ω : cf(x) > α} = {x ∈ Ω : f (x) < α/c} if c < 0.
Since both of these sets are measurable, cf is measurable for all c.
(iii) For all α ≥ 0, the set
{x ∈ Ω : |f(x)| > α} = {x ∈ Ω : f(x) > α} ∪ {x ∈ Ω : f (x) < −α}
is clearly measurable. If α < 0, then \left\{x ∈ Ω : |f (x)| > α\right\} = Ω ∈\mathcal{M}.
(iv) Let Ω_{1} = \left\{x ∈ Ω : |f (x)| < ∞\right\} and g =f|_{Ω_{1}}. Then Ω_{1} ∈ \mathcal{M} and g is measurable. Since the function \varphi(x) = x^{n} is continuous on \mathbb{R}, it follows from Remark 10.7.5 above that g^{n} = \varphi \circ g is measurable.
Also f^{n}(x) = ∞ if and only if |f(x)| = ∞, so (f^{n})^{−1}(\left\{∞\right\}) ∈ \mathcal{M}. On the other hand, (f^{n})^{−1}(\left\{−∞\right\}) is either empty or f^{−1}(\left\{−∞\right\}), and is therefore measurable. Finally, noting that every Borel set B lies in \mathbb{R}, we have
(f^{n})^{−1}(B) = (g^{n})^{−1}(B) ∈\mathcal{M},
so we conclude that f^{n} is measurable.
(v) Let Ω_{2} = \left\{x ∈ Ω : f (x) = 0\right\}, Ω_{3} = \left\{x ∈ Ω : f (x) = −∞\right\}, and Ω_{4} = Ω \setminus (Ω_{2} ∪ Ω_{3}). We shall prove that the functions
g = \frac{1}{f} \bigg|_{Ω_{2}} , h = \frac{1}{f} \bigg|_{Ω_{3}} , k = \frac{1}{f} \bigg|_{Ω_{4}}
are all measurable. By definition of 1/f, we have \left\{x ∈ Ω_{2} : g(x) < α\right\} = \varnothing for all α ∈ \mathbb{R}, so this set is measurable. Furthermore,
\left\{x ∈ Ω_{3} : h(x) < α\right\} = \begin{cases} Ω_{3}, & α > 0 \\ \varnothing, & α ≤ 0, \end{cases}
and in either case this set is also measurable. Finally we can write k = \varphi \circ f, where φ is the continuous function defined on \mathbb{R} \setminus \left\{0\right\} by φ(x) = 1/x. Hence k is measurable.
(vi) It is a simple matter to ascertain that the product f g restricted to any of the sets {x : f(x) = ∞}, {x : f(x) = −∞}, {x : g(x) = ∞}, {x : g(x) = −∞} is measurable. Therefore we may assume that f and g are real valued on all of Ω. Since
f g = \frac{1}{2} \left[(f + g)^{2} − f^{2} − g^{2}\right] ,
we conclude from (i), (ii), and (iv) that f g is measurable.