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This was taken out of Tao's book on Measure theory:

Let $f : \mathbb{R}^d \rightarrow [0, + \infty]$ be an unsigned measurable function. Show that the region $ \{ (x,t) \in \mathbb{R}^d \times \mathbb{R} : 0 \leq t \leq f(x) \}$ is a measurable set of $\mathbb{R}^{d+1}$

My attempt:

We know that the set $\{ x \in \mathbb{R}^d : t < f(x) \}$ is measurable (by definition)

Therefore $\bigcup_{t \in \mathbb{Q}} \{ x \in \mathbb{R}^d : t < f(x) \} \times \{t\}$ is measurable as it is the countable union of measurable sets.

This set is equal to $ \{ (x,t) \in \mathbb{R}^d \times \mathbb{Q} : 0 \leq t \leq f(x) \}$.

I was trying to find a density argument to conclude the desired result, but I am stuck, and this led me to think that this approach is fruitless. Any help?

proofromthebook
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1 Answers1

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I was also dealing with that question recently. It will be the best if someone can help verify my answer.

Since $f$ is measurable, there exists an increasing sequence of unsigned simple functions $\{f_n\}$ such that $\{f_n\}\to f$ pointwise. Explicitly, if $f_n=\sum_{m=1}^{M_n}c_{n,m}\textbf{1}_{E_{n,m}}$ for measurable subsets $E_{n,m}$, then $$S_n:=\{(x,t):0\leq t\leq f_n(x)\}=\bigcup_{m=1}^{M_n}E_{n,m}\times[0,c_{n,m}]$$ is a measurable set since product of measurable subsets is measurable (Exercise 1.2.22(ii)). Meanwhile, as $f_n\to f$ pointwise from below, we have $$\{(x,t)\in \mathbb{R}^d\times[0,+\infty]:0\leq t\leq f(x)\}=\bigcup_{n\in\mathbb{N}}S_n:=S.$$ Countable union preserves measurability, so $S$ is measurable.

Ivan So
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  • $S' = {(x, t)\in\mathbb R^d\times [0, +\infty]: 0\le t\le f(x)}}$ might not be equal to $\displaystyle S = \bigcup_{n\in\mathbb N} S_n$ since $(x, f(x))$ surely is in $S'$, but may not be in $S_n$ for any $n\in\mathbb N$ (if $f_n(x)$ is always strictly less than $f(x)$). – Alma Arjuna May 06 '23 at 18:37