Let $m$ denote the Lebesgue measure on $\mathbb{R}^n$. Let $\gamma:[a,b]\to \mathbb{R}^n$. Denote by $\Gamma=\{\gamma(t)|t\in[a,b]\}$, image of $\gamma$.
I know if $f$ is continuous then image is not of measure zero. For example take space filing curve(Peano curve?).
But what if I add a extra condition that function $\gamma$ is differentiable?
If $\gamma$ is rectifiable then its image has content zero.
Proof (for $n=2$; can be extended to $n>2$.)
Let $R\subset\Bbb R^2$ be a square such that $\Gamma\subset R$. Assume without oso of generality that its sides have length $2$. Divide $R$ into $4\,n^2$ equal squares of side $1/n^2$. Clasify them into four classes $A_n$, $B_n$, $C_n$ y $D_n$, each one containing $n^2$ squares, such that if $R_1$, $R_2$ are two different squares in the same group and $p_i\in R_i$, then $|p_1-p_2|\ge1/n$.
Suppose that for some $n$ $\Gamma$ has non empty intersection with at least $n\sqrt n$ squares in $A_n$. Then, considering a polygonal with a point in each of the squares, we have $$ \operatorname{length}(\gamma)\ge n\sqrt{n}\,\frac1n=\sqrt n\,. $$ Since $\gamma$ is rectifiable, there exists $n_A$ such that $\Gamma$ has non empty intersection with less than $n\,\sqrt{n}$ squares in $A_{n}$ for all $n\ge n_A$. The sum of the areas of all those squares is bounded by $$ n\,\sqrt{n}\,\frac1{n^2}=\frac{1}{\sqrt n}\,. $$ Repeating the argument with $B_n$, $C_n$ and $D_n$ we see that if $n\ge\max(n_A,n_B,n_C,n_D)$, entonces $\Gamma$ is contained ia set of squares of total area bounded by $4/\sqrt n$.