Let $A \subset \Bbb R^2$ be the graph $f: \Bbb R \to \Bbb R$ given by: $$ f(x) = \begin{cases} \text{sin($\frac{1}{x}$)}, & \text{if $x \in [-1,1]\setminus \{0\}$} \\[2ex] 5, & \text{if $x=0$} \end{cases} $$
Let $S = A \cup\{(0,y) \in \Bbb R^2 : -1 \le y \le 1\}$. Is $S$ Lebesgue measurable on the plane?
I see that $f(x)$ is continuous in the plane so I am thinking that is is measurable, but the point $(0,5)$ defined for the graph is throwing me off.
One of the strengths of measure theory is that single points shouldn't throw you off. If your function is equivalent to a continuous function almost everywhere, then its image will be measurable, since the only discontinuities occur on a set of measure zero.
as far as I know:
(1) continuous image of an interval is measurable, so $f([-1,0))$ and $f((0,1])$ are measurable sets
(2) points and intervals are measurable
hence $A=f([-1,0))\cup f((0,1]) \cup \{(0,5)\} $ is measurable and $S$ also.
