I am trying to solve this problem.
Let $f:[0,1] \to \mathbb{R}$ be a continuous function. For $t \in [0,1]$, define a line segment $S(t;f)$ by $$ S(t;f) := \{((1-\alpha)t + \alpha f(t), \alpha) \mid 0\le \alpha \le 1\}. $$ Let $\bar{S}(f) := \bigcup_{t\in[0,1]} S(t;f)$. Show that $\bar{S}(f)$ is Lebesgue measurable.
A set $A$ is defined to be Lebesgue measurable if $m_*(A) = m^*(A)$, where $$ m^*(A) = \inf \{\sum_{i=1}^\infty |I_i| \mid A \subset \bigcup_{i=1}^\infty I_i \} \qquad \text{(exterior measure)}\\ m_*(A) = |I| - m^*(I\setminus A) \qquad \text{(interior measure)}. $$ In the definition of $m^*(A)$, $I_i \subset \mathbb{R}^N$ is an interval, and in the definition of $m_*(A)$, $I\supset A$ is an interval.
I figured that if $f$ is monotone, then $\bar{S}(f)$ is a trapezoid, so that it is measurable (though I haven't figured how to prove it rigorously). But, if $f$ is not monotone, such as $f(x)=(x-0.5)^2$, then the shape of $\bar{S}(f)$ is strange, so I am guessing the geometric approach wouldn't work.
Could you give me any suggestion where to start from?
The mapping $\mathbb{R}^2\ni(t, \alpha)\mapsto ((1-\alpha)t+\alpha f(t), \alpha)\in \mathbb{R}^2$ is continuous and $\bar{S}(f)$ is the image of the compact set $[0,1]\times [0,1]$ under this mapping. Hence $\bar{S}(f)$ is compact, in particular measurable.