I am studying sards theorem and I want to prove "baby Sard's Theorem": Given $M$ and $N$ manifolds of dimension $m$ and $n$, respectively, $m < n$ and a smooth function $f:M\rightarrow N$, then $f(M)$ has measure zero in $N$.

I've seen and understood the "classical" proof where you compose with a projection and use that smooth maps between open sets of the same euclidean space send sets of measure zero to sets of measure zero, but I was trying another approach and got stuck.

I tried to prove that an smooth function $f: X\rightarrow\mathbb{R}^n$ with $X$ in $\mathbb{R}^m$ closed ball satisfies $f(X)$ has measure zero as follows: consider the line homotopy $F: (X\times [0, 1])\rightarrow \mathbb{R}^n$ given by $F(x, t) = tf(x) + (1 - t)(x, 0)$ and define $A = \{t\in [0, 1] | F_t(X) \;\mbox{has measure } 0 \}$.

Because $X$ is a closed ball of $\mathbb{R}^m$ it has measure zero in $\mathbb{R}^n$, since actually all of $\mathbb{R}^m\times 0$ has measure zero. That is, $0\in A$. Now I want to prove that $A$ is closed and open. Openess is easy since given any $\epsilon > 0$ and a cover of $F_t(X)$ of total volume less than $\epsilon$, we can make pull it back to $X\times [0, 1]$ to get a neighborhood $V$ of $X\times t$, then each point has a box around inside $V$, and by compactness picking a finite set of these boxes we have a that $X\times (t - \delta, t + \delta)$ is in $V$. $\delta$ has just to be smaller than the finite set of heights of the boxes.

But I can't prove it is closed. I pick $t_1,. t_2,...$ converging to $t$, such that $F_{t_k}(X)$ has measure zero, and I want to prove that eventually $F_{t_k}(X)$ is so close to $F_t(X)$ that the balls that cover it, actually cover $F_t(X)$, but I dont find a way to control these balls. As $t_k$ approach $t$ this balls may be getting very small in size altough more in number.

I tried to bound the amount of balls or the size of the radius they might have but with no succes. Does anyone see a way to fix this or know if it's doom to fail?

  • $\begingroup$ I don't know if this approach works, but you ought to use that the map is smooth (it doesn't hold for the continuous case). $\endgroup$ – Thomas Rot Oct 27 '16 at 7:23

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