# $f$ is measurable iff $f^+$ and $f^-$ are measurable (Çınlar Exercise 2.23)

Problem:

Discussion:

I've included some background (terminology) at the bottom of this post. I attempted this proof without seeking any assistance. I've since found this link and this link and they seem dissimilar enough from what I've done that I'm concerned that I've messed up.

Please note that at this point in the book, I am not privy to any information about the difference of two measurable functions being measurable.

My Attempt:

I guess the phrase $$\mathscr E$$-measurable'' implies that there are measurable spaces $$(E, \mathscr E)$$ and $$\big(\bar{\mathbb R}, \mathscr B(\bar{\mathbb R}) \big)$$, and functions $$f \colon E \to \bar{\mathbb R}, f^+ \colon E \to \bar{\mathbb R}$$ and $$f^- \colon E \to \bar{\mathbb R}$$.

I expect that we'll have to use the formula $$f = f^+ - f^-$$ in this proof.

Suppose $$f$$ is $$\mathscr E$$-measurable. So for example, for every $$r \in \mathbb R$$, $$f^{-1}[-\infty, r] \in \mathscr E$$. First we try to show that $$f^+$$ is $$\mathscr E$$-measurable, so we want to show that for any $$r \in \mathbb R$$, $$(f \lor 0)^{-1}[-\infty, r] \in \mathscr E$$. Consider two cases.

Case 1: $$r < 0$$. Then $$(f \lor 0)^{-1}[-\infty, r] = \emptyset \in \mathscr E$$ because $$f^+$$ sends nothing to negative real numbers, and the empty set must be in $$\mathscr E$$.

Case 2: $$r \geq 0$$. Then break up $$[-\infty, r]$$ into the disjoint union $$[-\infty, 0) \cup \{ 0 \} \cup (0, r]$$ and we see that

\begin{align*} (f \lor 0)^{-1}[-\infty, r] &= (f \lor 0)^{-1}[-\infty, 0) \cup \{ 0 \} \cup (0, r]\\ &= (f \lor 0)^{-1}[-\infty, 0) \cup (f \lor 0)^{-1} \{ 0 \} \cup (f \lor 0)^{-1} (0, r]\\ &= \emptyset \cup f^{-1}[-\infty, 0] \cup f^{-1}(0, r]\\ &= f^{-1}[-\infty, 0] \cup f^{-1}(0, r]. \end{align*}

Since $$f$$ is $$\mathscr E$$-measurable and $$[-\infty, 0]$$ and $$(0, r]$$ are both Borel sets, i.e. they're both in $$\mathscr B(\bar{\mathbb R})$$, then both preimages must be in $$\mathscr E$$, so the union of both preimages must also be in $$\mathscr E$$, and we conclude that $$f^+$$ is $$\mathscr E$$-measurable.

Second we try to show that $$f^-$$ is $$\mathscr E$$-measurable, so we want to show that for any $$r \in \mathbb R$$, $$\big(-(f \land 0)\big)^{-1}[-\infty, r] \in \mathscr E$$. Again consider two cases.

Case 1: $$r < 0$$. Then $$\big(-(f \land 0)\big)^{-1}[-\infty, r] = \emptyset \in \mathscr E$$ because $$f^-$$ sends nothing to negative real numbers, and the empty set must be in $$\mathscr E$$.

Case 2: $$r \geq 0$$. Then break up $$[-\infty, r]$$ into the disjoint union $$[-\infty, 0) \cup \{ 0 \} \cup (0, r]$$ and we see that

\begin{align*} \big(-(f \land 0)\big)^{-1}[-\infty, r] &= \big(-(f \land 0)\big)^{-1}[-\infty, 0) \cup \{ 0 \} \cup (0, r]\\ &= \big(-(f \land 0)\big)^{-1}[-\infty, 0) \cup \big(-(f \land 0)\big)^{-1} \{ 0 \} \cup \big(-(f \land 0)\big)^{-1} (0, r]\\ &= \emptyset \cup f^{-1} [0, \infty] \cup f^{-1}[-r, 0)\\ &= f^{-1} [0, \infty] \cup f^{-1}[-r, 0) \end{align*}

and since $$f$$ is $$\mathscr E$$-measurable and each of the intervals $$[0, \infty]$$ and $$[-r, 0)$$ is a Borel set, i.e. they're each in $$\mathscr B(\bar{\mathbb R})$$, then we must have $$f^{-1}[0, \infty] \in \mathscr E$$ and $$f^{-1}[-r, 0) \in \mathscr E$$, so the union of the preimages must be in $$\mathscr E$$, and we conclude that $$f^-$$ is $$\mathscr E$$-measurable.

Now for the converse, suppose $$f^+$$ and $$f^-$$ are both $$\mathscr E$$-measurable. We want to show that $$f$$ is $$\mathscr E$$-measurable, which could be done by showing that for any $$r \in \mathbb R, f^{-1}[-\infty, r] \in \mathscr E$$. We will apply the decomposition formula $$f = f^+ - f^-$$. Consider two cases.

Case 1: $$r < 0$$. Then $$f = -f^-$$, so $$f^{-1}[-\infty, r] = (-f^-)^{-1}[-\infty, r] = (f^-)^{-1}[-r, \infty] \in \mathscr E$$ since $$f^-$$ is $$\mathscr E$$-measurable and $$[-r, \infty]$$ is a Borel set, i.e. it's in $$\mathscr B(\bar{\mathbb R})$$.

Case 2: $$r \geq 0$$. Then break up $$[-\infty, r]$$ into the disjoint union $$[-\infty, 0) \cup \{ 0 \} \cup (0, r]$$ and we see that

\begin{align*} f^{-1}[-\infty, r] &= f^{-1}[-\infty, 0) \cup \{ 0 \} \cup (0, r]\\ &= f^{-1}[-\infty, 0) \cup f^{-1} \{ 0 \} \cup f^{-1} (0, r]\\ &= (-f^-)^{-1}[-\infty, 0) \cup (-f^-)^{-1} \{ 0 \} \cup (f^+)^{-1} (0, r]\\ &= (f^-)^{-1}(0, \infty] \cup (f^-)^{-1} \{ 0 \} \cup (f^+)^{-1} (0, r]. \end{align*}

Each of the components of the union must lie in $$\mathscr E$$ because the functions $$f^-$$ and $$f^+$$ are $$\mathscr E$$-measurable and the sets $$(0, \infty], \{ 0 \}$$ and $$(0, r]$$ are Borel sets, i.e. they're in $$\mathscr B(\bar{\mathbb R})$$, and therefore the union lies in $$\mathscr E$$, so we conclude that $$f$$ is $$\mathscr E$$-measurable.

Background:

Here I'm pasting some terminology from the book.

Your brute force approach seems correct. But such an approach will become unwieldy for more complicated functions. To reduce the number of cases you have to check, the systematic approach to these is to use the following lemma:

Let $$(X, M)$$ be a measure space. Let $$f : X \to \overline{\mathbb{R}}$$ and $$Y = f^{-1}(\mathbb{R})$$. Then $$f$$ is measurable if and only if $$f^{-1}(\{-\infty\}) \in M, f^{-1}(\{\infty\}) \in M$$, and $$f$$ is measurable on $$Y$$ (Folland chapter 2 exercise 2).

For example, addition $$+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$ is $$B(\mathbb{R} \times \mathbb{R}) \to B(\mathbb{R})$$ measurable since it is continuous. Then you would use the lemma to show that addition $$+ : \overline{\mathbb{R}} \times \overline{\mathbb{R}} \to \overline{\mathbb{R}}$$ is measurable, where you define $$\infty - \infty$$ arbitrarily. Then since composition of measurable functions is measurable, this (along with $$B(\mathbb{R} \times \mathbb{R}) = B(\mathbb{R}) \otimes B(\mathbb{R})$$) implies that the sum of two measurable functions is measurable.

$$\sup$$ and $$\inf$$ of a sequence $$f_n$$ of measurable functions can be shown to be measurable directly.

Combining the two above results proves your theorem.

• Thanks. This seems like a much more sophisticated way of dealing with the issue. I don't fully get it but I'll make a note of it and try to make sense of it. Jan 16 at 18:51
• All of it is elementary to prove, the hardest one being $$X, Y \text{ second-countable } \implies B(X \times Y) = B(X) \otimes B(Y),$$ for which you need to know about the product topology and product sigma algebra to prove. But it isn't much more work than you have already done just for this one problem, and the results are applicable everywhere. Jan 16 at 20:03