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$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).

Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this MathSE post (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).

Taking into accounts the comments: here is a setting encompassing both the cases when $f$ is locally Lipschitz, and when $f$ is differentiable.

Suppose that for every $x\in\mathbf{R}^n$, we have $$(*)\qquad F_f(x)=\limsup_{y\to x,\;y\neq x}\frac{\|f(y)-f(x)\|}{\|y-x\|}<\infty.$$

Define, for $p$ positive integer $$X_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\|y-x\|\le 1/p \Rightarrow \|f(y)-f(x)\|\le p\|y-x\|\}.$$ Then $\mathbf{R}^n$ is the (countable) union of all $X_p$, and $X_p$ is a countable union of subsets $X_{p,i}$ of diameter $\le 1/p$. And $f$ is $p$-Lipschitz on $X_{p,i}$ (and also on its closure, in case one wishes to get closed subsets).

So the result indeed follows, not of the Lipschitz case as strictly said, but of the same statement replacing $\mathbf{R}^n$ with a subset of $\mathbf{R}^n$ with the restriction of the Euclidean distance (namely: for $n<m$ and $Y$ subset of $\mathbf{R}^n$, every Lipschitz function $Y\to\mathbf{R}^m$ has image of measure zero). The argument for the latter seems unchanged.

PS: for a reference, it is mentioned by @Koch that Lemma 7.25 in Rudin's Real and complex analysis (initially published in 1966) does all the job: it asserts that any map $f:\mathbf{R}^m\to\mathbf{R}$ satisfying $(*)$ maps measure zero subsets to measure zero subsets. The proof given here actually seems to roughly be the same as the one written (concisely) in Rudin's book.

$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).

Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this MathSE post (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).

Taking into accounts the comments: here is a setting encompassing both the cases when $f$ is locally Lipschitz, and when $f$ is differentiable.

Suppose that for every $x\in\mathbf{R}^n$, we have $$(*)\qquad F_f(x)=\limsup_{y\to x,\;y\neq x}\frac{\|f(y)-f(x)\|}{\|y-x\|}<\infty.$$

Define, for $p$ positive integer $$X_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\|y-x\|\le 1/p \Rightarrow \|f(y)-f(x)\|\le p\|y-x\|\}.$$ Then $\mathbf{R}^n$ is the (countable) union of all $X_p$, and $X_p$ is a countable union of subsets $X_{p,i}$ of diameter $\le 1/p$. And $f$ is $p$-Lipschitz on $X_{p,i}$ (and also on its closure, in case one wishes to get closed subsets).

So the result indeed follows, not of the Lipschitz case as strictly said, but of the same statement replacing $\mathbf{R}^n$ with a subset of $\mathbf{R}^n$ with the restriction of the Euclidean distance (namely: for $n<m$ and $Y$ subset of $\mathbf{R}^n$, every Lipschitz function $Y\to\mathbf{R}^m$ has image of measure zero). The argument for the latter seems unchanged.

PS: for a reference, it is mentioned by @Kosh that Lemma 7.25 in Rudin's Real and complex analysis (initially published in 1966) does all the job: it asserts that any map $f:\mathbf{R}^m\to\mathbf{R}$ satisfying $(*)$ maps measure zero subsets to measure zero subsets. The proof given here actually seems to roughly be the same as the one written (concisely) in Rudin's book.

$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).

Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this MathSE post (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).

Taking into accounts the comments: here is a setting encompassing both the cases when $f$ is locally Lipschitz, and when $f$ is differentiable.

Suppose that for every $x\in\mathbf{R}^n$, we have $$F_f(x)=\limsup_{y\to x,\;y\neq x}\frac{\|f(y)-f(x)\|}{\|y-x\|}<\infty.$$

Define, for $p$ positive integer $$X_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\|y-x\|\le 1/p \Rightarrow \|f(y)-f(x)\|\le p\|y-x\|\}.$$ Then $\mathbf{R}^n$ is the (countable) union of all $X_p$, and $X_p$ is a countable union of subsets $X_{p,i}$ of diameter $\le 1/p$. And $f$ is $p$-Lipschitz on $X_{p,i}$ (and also on its closure, in case one wishes to get closed subsets).

So the result indeed follows, not of the Lipschitz case as strictly said, but of the same statement replacing $\mathbf{R}^n$ with a subset of $\mathbf{R}^n$ with the restriction of the Euclidean distance (namely: for $n<m$ and $Y$ subset of $\mathbf{R}^n$, every Lipschitz function $Y\to\mathbf{R}^m$ has image of measure zero). The argument for the latter seems unchanged.

$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).

Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this MathSE post (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).

Taking into accounts the comments: here is a setting encompassing both the cases when $f$ is locally Lipschitz, and when $f$ is differentiable.

Suppose that for every $x\in\mathbf{R}^n$, we have $$(*)\qquad F_f(x)=\limsup_{y\to x,\;y\neq x}\frac{\|f(y)-f(x)\|}{\|y-x\|}<\infty.$$

Define, for $p$ positive integer $$X_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\|y-x\|\le 1/p \Rightarrow \|f(y)-f(x)\|\le p\|y-x\|\}.$$ Then $\mathbf{R}^n$ is the (countable) union of all $X_p$, and $X_p$ is a countable union of subsets $X_{p,i}$ of diameter $\le 1/p$. And $f$ is $p$-Lipschitz on $X_{p,i}$ (and also on its closure, in case one wishes to get closed subsets).

So the result indeed follows, not of the Lipschitz case as strictly said, but of the same statement replacing $\mathbf{R}^n$ with a subset of $\mathbf{R}^n$ with the restriction of the Euclidean distance (namely: for $n<m$ and $Y$ subset of $\mathbf{R}^n$, every Lipschitz function $Y\to\mathbf{R}^m$ has image of measure zero). The argument for the latter seems unchanged.

PS: for a reference, it is mentioned by @Koch that Lemma 7.25 in Rudin's Real and complex analysis (initially published in 1966) does all the job: it asserts that any map $f:\mathbf{R}^m\to\mathbf{R}$ satisfying $(*)$ maps measure zero subsets to measure zero subsets. The proof given here actually seems to roughly be the same as the one written (concisely) in Rudin's book.

$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).

Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this MathSE post (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).

$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).

Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this MathSE post (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).

Taking into accounts the comments: here is a setting encompassing both the cases when $f$ is locally Lipschitz, and when $f$ is differentiable.

Suppose that for every $x\in\mathbf{R}^n$, we have $$F_f(x)=\limsup_{y\to x,\;y\neq x}\frac{\|f(y)-f(x)\|}{\|y-x\|}<\infty.$$

Define, for $p$ positive integer $$X_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\|y-x\|\le 1/p \Rightarrow \|f(y)-f(x)\|\le p\|y-x\|\}.$$ Then $\mathbf{R}^n$ is the (countable) union of all $X_p$, and $X_p$ is a countable union of subsets $X_{p,i}$ of diameter $\le 1/p$. And $f$ is $p$-Lipschitz on $X_{p,i}$ (and also on its closure, in case one wishes to get closed subsets).

So the result indeed follows, not of the Lipschitz case as strictly said, but of the same statement replacing $\mathbf{R}^n$ with a subset of $\mathbf{R}^n$ with the restriction of the Euclidean distance (namely: for $n<m$ and $Y$ subset of $\mathbf{R}^n$, every Lipschitz function $Y\to\mathbf{R}^m$ has image of measure zero). The argument for the latter seems unchanged.