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We say that $f:[0,1]\to \mathbb{R}$ satisfies Lusin's condition (N) provided $$m(f(B))=0 \quad\mbox{whenever}\quad B\subseteq [0,1] \mbox{ with }m(B)=0$$ where $m$ stands for the Lebesgue measure on $\mathbb{R}$.

I found in here the following definition.

We say that $f:[0,1]\to X$ satisfies Lusin's condition (N) provided $$\mathcal{H}^1(f(B))=0 \quad\mbox{whenever}\quad B\subseteq [0,1] \mbox{ with }m(B)=0$$ where $X$ is a metric space and $\mathcal{H}^1$ stands for 1-dimensional Hausdorff measure.

What comes to my mind is the possible formulation of condition (N) for the function $f:[0,1]\to X$ but this time our $X$ is a Hausdorff locally convex topological vector space. I don't know if such formulation exists. If such formulation exists, I would be greatful if you can provide me such definition.

Juniven Acapulco
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1 Answers1

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Lusin's condition (N) deals exclusively with the measure space structure. A measurable map $f:X\to Y$ between measure spaces $X=(X,\Sigma_X,\mu_X)$ and $Y=(Y,\Sigma_Y,\mu_Y)$ satisfies the condition (N) if $$ \mu_X(B)=0\implies \mu_Y(f(B))=0\tag{N}$$

The rest concerns not (N) itself, but rather the availability of measures on our spaces. On any metric space we have the family of Hausdorff measures $\mathcal H^s$. So one can choose (as the authors you quoted did) to use $\mathcal H^1$ on the target space when the domain is one-dimensions. They would probably use $\mathcal H^n$ on the target space if the domain was $n$-dimensional.

If your locally convex space is metrizable, you can use a metric to define $\mathcal H^1$ and hence, the property (N). However, the property will depend on the choice of the metric, even if it is required to be translation-invariant. Indeed, if $d$ is one such metric, than $d^{\alpha}$ is another (for any $\alpha\in (0,1)$). The Hausdorff measure $\mathcal H^1$ with respect to $d^\alpha$ is equivalent to the Hausdorff measure $\mathcal H^{\alpha}$ with respect to $d$. This measure is quite different from $\mathcal H^{1}$ with respect to $d$; the ideals of null sets are not the same.

Conclusion: If the target space lacks a canonical choice of a measure, there is no canonical notion of condition (N) for maps into it.

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