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Union Measurability of the union of cut locusloci along a curve

Let $(M,g)$ be a Riemannian symmetric space and $f:M\to\mathbb{R}$$\alpha(s)$ be a non-negative integrable function. Fix a geodesic $\alpha(t)$. Define $$ U(t):=\cup_{s\in[0,t]}{\rm Cut}(\alpha(s)) $$$$ U(t)=\cup_{s\in[0,t]}{\rm Cut}(\alpha(s)) $$ as the union of the cut locusloci ${\rm Cut}(\alpha(s))$ of $\alpha(s)$,along the curve $s\in[0,t]$$\alpha$. Questions:

(1) IsI am curious whether the set $U(t)$ ais measurable set?

(2) Is the following quantity finite: $$ {\lim\sup}_{t\to 0}\frac{\int_{U(t)}f(x) {\rm dvol}(x)}{t}. $$or not. One may assume $t$ is small whenever necessary.

Union of the cut locus

Let $(M,g)$ be a Riemannian symmetric space and $f:M\to\mathbb{R}$ be a non-negative integrable function. Fix a geodesic $\alpha(t)$. Define $$ U(t):=\cup_{s\in[0,t]}{\rm Cut}(\alpha(s)) $$ as the union of the cut locus ${\rm Cut}(\alpha(s))$ of $\alpha(s)$, $s\in[0,t]$. Questions:

(1) Is $U(t)$ a measurable set?

(2) Is the following quantity finite: $$ {\lim\sup}_{t\to 0}\frac{\int_{U(t)}f(x) {\rm dvol}(x)}{t}. $$

Measurability of the union of cut loci along a curve

Let $(M,g)$ be a Riemannian symmetric space and $\alpha(s)$ be a geodesic. Define $$ U(t)=\cup_{s\in[0,t]}{\rm Cut}(\alpha(s)) $$ as the union of the cut loci ${\rm Cut}(\alpha(s))$ along the curve $\alpha$. I am curious whether the set $U(t)$ is measurable or not. One may assume $t$ is small whenever necessary.

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Union of the cut locus

Let $(M,g)$ be a Riemannian symmetric space and $f:M\to\mathbb{R}$ be a non-negative integrable function. Fix a geodesic $\alpha(t)$. Define $$ U(t):=\cup_{s\in[0,t]}{\rm Cut}(\alpha(s)) $$ as the union of the cut locus ${\rm Cut}(\alpha(s))$ of $\alpha(s)$, $s\in[0,t]$. Questions:

(1) Is $U(t)$ a measurable set?

(2) Is the following quantity finite: $$ {\lim\sup}_{t\to 0}\frac{\int_{U(t)}f(x) {\rm dvol}(x)}{t}. $$