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A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see

Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Math. J. 6 (1999), no. 3, 201–212.

Edit: As requested, here's a summary of the proof.

The authors prove that an arbitrary union of (closed, nondegenerate) $n$-simplices $\{ S_t \}_{t \in T}$ in $\mathbb{R}^n$ is Lebesgue measurable. First a preliminary definition:

A bounded set $X \subset \mathbb{R}^n$ is said to be $\alpha$-regular, for $\alpha$ a positive real number, if $\lambda(X) \geq \alpha \lambda(V(X))$, where $V(X)$ is a closed ball with minimal diameter for which the inclusion $X \subset V(X)$ holds.

Observe that each $n$-simplex is $\alpha$-regular for some $\alpha \in (0,1]$. Thus

$$ \bigcup_{t \in T} S_t = \bigcup_{m=1}^\infty \ \bigcup \{ S_t \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular} \}. $$

So in order to show that $\cup_t S_t$ is Lebesgue measurable, it suffices to show that $X_m = \cup \{ S_t \colon S_t \text{ is } \frac{1}{m}\text{-regular} \}$ is Lebesgue measurable for all $m \in \mathbb{Z}_{>0}$. Towards this end, given $x \in S_t$ and $c \in (0,1)$, let $S_t(x,c)$ denote the image of $S_t$ under the map $y \mapsto x + c(y-x)$. Then

$$ \mathcal{F}_m = \{ S_t(x,c) \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular}, x \in S_t, c \in (0,1) \} $$

is a Vitali covering of $X_m$. The Vitali covering theorem now takes us home: the countable subcollection $\mathcal{F}_m^\ast \subset \mathcal{F}_m$ produced by the theorem has a Lebesgue measurable union $\cup \mathcal{F}_m^\ast$, which also satisfies

$$ \bigcup \mathcal{F}_m^\ast \subset X_m \quad\text{and}\quad \lambda(X_m \backslash \bigcup \mathcal{F}_m^\ast) = 0. $$

A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see

Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Math. J. 6 (1999), no. 3, 201–212.

Edit: As requested, here's a summary of the proof.

The authors prove that an arbitrary union of (closed, nondegenerate) $n$-simplices $\{ S_t \}_{t \in T}$ in $\mathbb{R}^n$ is Lebesgue measurable. First a preliminary definition:

A bounded set $X \subset \mathbb{R}^n$ is said to be $\alpha$-regular, for $\alpha$ a positive real number, if $\lambda(X) \geq \alpha \lambda(V(X))$, where $V(X)$ is a closed ball with minimal diameter for which the inclusion $X \subset V(X)$ holds.

Observe that an $n$-simplex is $\alpha$-regular for some $\alpha \in (0,1]$. Thus

$$ \bigcup_{t \in T} S_t = \bigcup_{m=1}^\infty \ \bigcup \{ S_t \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular} \}. $$

So in order to show that $\cup_t S_t$ is Lebesgue measurable, it suffices to show that $X_m = \cup \{ S_t \colon S_t \text{ is } \frac{1}{m}\text{-regular} \}$ is Lebesgue measurable for all $m \in \mathbb{Z}_{>0}$. Towards this end, given $x \in S_t$ and $c \in (0,1)$, let $S_t(x,c)$ denote the image of $S_t$ under the map $y \mapsto x + c(y-x)$. Then

$$ \mathcal{F}_m = \{ S_t(x,c) \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular}, x \in S_t, c \in (0,1) \} $$

is a Vitali covering of $X_m$. The Vitali covering theorem now takes us home: the countable subcollection $\mathcal{F}_m^\ast \subset \mathcal{F}_m$ produced by the theorem has a Lebesgue measurable union $\cup \mathcal{F}_m^\ast$, which also satisfies

$$ \bigcup \mathcal{F}_m^\ast \subset X_m \quad\text{and}\quad \lambda(X_m \backslash \bigcup \mathcal{F}_m^\ast) = 0. $$

A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see

Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Math. J. 6 (1999), no. 3, 201–212.

Edit: As requested, here's a summary of the proof.

The authors prove that an arbitrary union of (closed, nondegenerate) $n$-simplices $\{ S_t \}_{t \in T}$ in $\mathbb{R}^n$ is Lebesgue measurable. First a preliminary definition:

A bounded set $X \subset \mathbb{R}^n$ is said to be $\alpha$-regular, for $\alpha$ a positive real number, if $\lambda(X) \geq \alpha \lambda(V(X))$, where $V(X)$ is a closed ball with minimal diameter for which the inclusion $X \subset V(X)$ holds.

Observe that each $n$-simplex is $\alpha$-regular for some $\alpha \in (0,1]$. Thus

$$ \bigcup_{t \in T} S_t = \bigcup_{m=1}^\infty \ \bigcup \{ S_t \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular} \}. $$

So in order to show that $\cup_t S_t$ is Lebesgue measurable, it suffices to show that $X_m = \cup \{ S_t \colon S_t \text{ is } \frac{1}{m}\text{-regular} \}$ is Lebesgue measurable for all $m \in \mathbb{Z}_{>0}$. Towards this end, given $x \in S_t$ and $c \in (0,1)$, let $S_t(x,c)$ denote the image of $S_t$ under the map $y \mapsto x + c(y-x)$. Then

$$ \mathcal{F}_m = \{ S_t(x,c) \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular}, x \in S_t, c \in (0,1) \} $$

is a Vitali covering of $X_m$. The Vitali covering theorem now takes us home: the countable subcollection $\mathcal{F}_m^\ast \subset \mathcal{F}_m$ the theorem produces is has a Lebesgue measurable union $\cup \mathcal{F}_m^\ast$, which also satisfies

$$ \bigcup \mathcal{F}_m^\ast \subset X_m \quad\text{and}\quad \lambda(X_m \backslash \mathcal{F}_m) = 0. $$

A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see

Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Math. J. 6 (1999), no. 3, 201–212.

Edit: As requested, here's a summary of the proof.

The authors prove that an arbitrary union of (closed, nondegenerate) $n$-simplices $\{ S_t \}_{t \in T}$ in $\mathbb{R}^n$ is Lebesgue measurable. First a preliminary definition:

A bounded set $X \subset \mathbb{R}^n$ is said to be $\alpha$-regular, for $\alpha$ a positive real number, if $\lambda(X) \geq \alpha \lambda(V(X))$, where $V(X)$ is a closed ball with minimal diameter for which the inclusion $X \subset V(X)$ holds.

Observe that each $n$-simplex is $\alpha$-regular for some $\alpha \in (0,1]$. Thus

$$ \bigcup_{t \in T} S_t = \bigcup_{m=1}^\infty \ \bigcup \{ S_t \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular} \}. $$

So in order to show that $\cup_t S_t$ is Lebesgue measurable, it suffices to show that $X_m = \cup \{ S_t \colon S_t \text{ is } \frac{1}{m}\text{-regular} \}$ is Lebesgue measurable for all $m \in \mathbb{Z}_{>0}$. Towards this end, given $x \in S_t$ and $c \in (0,1)$, let $S_t(x,c)$ denote the image of $S_t$ under the map $y \mapsto x + c(y-x)$. Then

$$ \mathcal{F}_m = \{ S_t(x,c) \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular}, x \in S_t, c \in (0,1) \} $$

is a Vitali covering of $X_m$. The Vitali covering theorem now takes us home: the countable subcollection $\mathcal{F}_m^\ast \subset \mathcal{F}_m$ produced by the theorem has a Lebesgue measurable union $\cup \mathcal{F}_m^\ast$, which also satisfies

$$ \bigcup \mathcal{F}_m^\ast \subset X_m \quad\text{and}\quad \lambda(X_m \backslash \bigcup \mathcal{F}_m^\ast) = 0. $$

A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see

Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Math. J. 6 (1999), no. 3, 201–212.

A google search reveals that an arbitrary union of (nondegenerate) convex sets is Lebesgue measurable: see

Balcerzak and Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Math. J. 6 (1999), no. 3, 201–212.

Edit: As requested, here's a summary of the proof.

The authors prove that an arbitrary union of (closed, nondegenerate) $n$-simplices $\{ S_t \}_{t \in T}$ in $\mathbb{R}^n$ is Lebesgue measurable. First a preliminary definition:

A bounded set $X \subset \mathbb{R}^n$ is said to be $\alpha$-regular, for $\alpha$ a positive real number, if $\lambda(X) \geq \alpha \lambda(V(X))$, where $V(X)$ is a closed ball with minimal diameter for which the inclusion $X \subset V(X)$ holds.

Observe that each $n$-simplex is $\alpha$-regular for some $\alpha \in (0,1]$. Thus

$$ \bigcup_{t \in T} S_t = \bigcup_{m=1}^\infty \ \bigcup \{ S_t \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular} \}. $$

So in order to show that $\cup_t S_t$ is Lebesgue measurable, it suffices to show that $X_m = \cup \{ S_t \colon S_t \text{ is } \frac{1}{m}\text{-regular} \}$ is Lebesgue measurable for all $m \in \mathbb{Z}_{>0}$. Towards this end, given $x \in S_t$ and $c \in (0,1)$, let $S_t(x,c)$ denote the image of $S_t$ under the map $y \mapsto x + c(y-x)$. Then

$$ \mathcal{F}_m = \{ S_t(x,c) \colon S_t \text{ is } \textstyle{\frac{1}{m}}\text{-regular}, x \in S_t, c \in (0,1) \} $$

is a Vitali covering of $X_m$. The Vitali covering theorem now takes us home: the countable subcollection $\mathcal{F}_m^\ast \subset \mathcal{F}_m$ the theorem produces is has a Lebesgue measurable union $\cup \mathcal{F}_m^\ast$, which also satisfies

$$ \bigcup \mathcal{F}_m^\ast \subset X_m \quad\text{and}\quad \lambda(X_m \backslash \mathcal{F}_m) = 0. $$