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Given that $x\left(t\right)$ is locally absolutely continuous, and $\dot{x}=f\left(x,t\right)$ exists almost everywhere, is it possible to place a condition on $f\left(x,t\right)$ to allow us to show that $x\left(t\right)$ is uniformly continuous? Perhaps that $\dot{x}=f\left(x,t\right)$ is locally uniformly bounded in t? But I'm not sure how to prove this and have not been able to find any specific resources that point to this fact.

I understand that absolutely continuous functions are by nature uniformly continuous, and if $x\left(t\right)$ is differentiable everywhere and $\dot{x}=f\left(x,t\right)$ is bounded everywhere, the $x\left(t\right)$ is UC. However, I am at a loss as to how to extend UC to these measure variations.

Thanks for any help!

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