Showing region between $x$ axis and graph is in ${\cal{B}}(\mathbb{R^2})$ Let $(\mathbb{R^2},{\cal{B}}(\mathbb{R^2}),\lambda^2)$ be a measure space where $\lambda^2$ is the two dimensional Lebesgue measure.
Let $u:\mathbb{R}\to [0,\infty]$ be a Borel measurable function. How do I show that $S[u]:=\{(x,y):0\le y \le u(x)\}$ is in ${\cal{B}}(\mathbb{R^2})$?
I was given the hint to approximate $u$ by an increasing sequence of simple functions $(f_j)$ that converges pointwise to $u$. Then $S[u]=\bigcup_j S[f_j]$ and since $S[f_j]\in {\cal{B}}(\mathbb{R^2})$ we conclude that $S[u]$ is also in ${\cal{B}}(\mathbb{R^2})$. 
However, how do we see that $S[u]=\bigcup_j S[f_j]$ actually holds?
 A: Edit: As pointed out by satokun in the comments, my answer is only correct if $S[u]$ would have been defined as $S[u] = \{(x,y) : 0 \leq y < u(x)\}$ instead of $S[u] = \{(x,y) : 0 \leq y \leq u(x)\}$. This means that either the hint is misleading or that there is a misprint in the exercise. There is also a nice blog post about this problem, found here (this post was also found by satokun). According to this post, we have the following possibilites:


*

*The hint in the problem is correct. This means that there is probably a misprint in the definition of $S[u]$ (as mentioned above). If this is the case, then my answer remains correct.

*The definition of $S[u]$ is correct. In this case the blog post shows that the hint is probably wrong. Furthermore, we don't know if $S[u]$ is considered as a subset of $\mathbb{R}^2$ or $\bar{\mathbb{R}} \times \bar{\mathbb{R}}$. Since the underlying set of the measure space is $\mathbb{R}^2$ and we have to show that $S[u] \in \mathcal{B}(\mathbb{R}^2)$, we can assume that it is meant a subset of $\mathbb{R}^2$, and in this case the correct answer is provided by the second-last paragraph of the blog post. However, if $S[u]$ is meant as a subset of $\bar{\mathbb{R}} \times \bar{\mathbb{R}}$, then in general $S[u] \notin \mathcal{B}(\mathbb{R}^2)$ without any further assumptions on $u$.

*The defintion of $u$ is not correct and should have been $u: \mathbb{R} \rightarrow [0, \infty)$ instead of $u: \mathbb{R} \rightarrow [0, \infty]$. Then $S[u] \subseteq \mathbb{R}^2$, and if the definiton of $S[u]$ is correct, the answer is given again by second last paragraph of the blog post. Otherwise (i.e. when there should have been $0 \leq y < u(x)$ instead of $0 \leq y \leq u(x)$ in the definiton of $S[u]$), the answer is given below.


Maybe someone else sees this problem and can give a better clarification about it.
Answer: Approximating $u$ by an increasing sequence of simple functions $(f_j)$ means that
\begin{equation}
f_j(x) \leq u(x)
\end{equation}
for all $x \in \mathbb{R}$ and $j \in \mathbb{N}$ (see here for the standard construction of the functions $f_j$). If we write for $x \in \mathbb{R}$
\begin{equation}
S[u]_x := \{x\} \times \{y \in \mathbb{R} : 0 \leq y \leq u(x) \} \subseteq S[u] 
\end{equation}
then it follows that
\begin{equation}
S[f_j]_x \subseteq S[u]_x 
\end{equation}
for all $x \in \mathbb{R}$ and $j \in \mathbb{N}$. Because ($f_j$) converges pointwise to $u$, we have
\begin{equation}
\bigcup_{j = 0}^{\infty}S[f_j]_x = S[u]_x 
\end{equation}
and because
\begin{equation}
\bigcup_{x \in \mathbb{R}} S[u]_x = S[u], \,\bigcup_{x \in \mathbb{R}} S[f_j]_x = S[f_j]
\end{equation}
we get that
\begin{equation}
S[u] = \bigcup_{x \in \mathbb{R}} S[u]_x = \bigcup_{x \in \mathbb{R}} \big(\bigcup_{j = 0}^{\infty}S[f_j]_x\big) = \bigcup_{j = 0}^{\infty} \big(\bigcup_{x \in \mathbb{R}} S[f_j]_x\big) = \bigcup_{j = 0}^{\infty}S[f_j] 
\end{equation}
So the equality holds indeed, and your conclusion is right.
