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Let $f: [a,b] \rightarrow ℝ^2$ be lipschitz continuous. Show that $\{f(t): t\in[a,b]\}$ is a null set.

I've already shown that the image of every null set for lipschitz continuous functions is also a null set. Can I extend this for an interval $[a,b]$? Can anyone give me a hint?

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Partition $ [a, b] $ into $ n $ closed intervals $ U_1, U_2, \ldots, U_n $ of equal measure in the obvious way. Since $ f $ is Lipschitz, it maps each of the $ U_k $ into an open ball $ B_k $ of radius $ (a-b)K/n $ in $ \mathbb R^2 $, where $ K $ is a Lipschitz constant of $ f $. The total measure of $ B_k $ is

$$ n \cdot \pi \cdot \frac{(a-b)^2 K^2}{n^2} = \pi \cdot \frac{(a-b)^2 K^2}{n} $$

which is an upper bound for the measure of the image of $ f $, since the $ B_k $ are an open cover of $ \textrm{Im} f $. Since $ n \in \mathbb N $ was arbitrary, let $ n \to \infty $ to obtain the result.

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