$L$ is submanifold of $\mathbb{R}^n$ with $\dim L\leq n-2$. [duplicate]

Let $x\in \mathbb R^n$. Define $F: (L\setminus \{x\}) \times \mathbb R\to \mathbb R^n$, $$ F(y, t) = tx + (1-t)y.$$

Clearly this map is smooth. Since $(L\setminus \{x\} \times \mathbb R)$ has dimension strictly less then $n= \dim \mathbb R^n$, the image of $F$ is of measure zero.

That means (i) when $x\notin L$, the union of the sets of lines in $\mathbb R^n$ passing through $x$ and $L$ is of measure zero in $\mathbb R^n$, and when (ii) $x\in L$, the union of sets of lines joining $x$ to some $y\in L$ in $\mathbb R^n$ is of measure zero.

In general, there is a more natural way to talk about measure zero in this setting: for each $x\in \mathbb R^n$, let $\mathbb{RP}^{n-1}_x$ be the set of lines in $\mathbb R^n$ passing through $x$. Then $\mathbb {RP}^{n-1}_x$ is a smooth manifold of dimension $n-1$ and "sets of measure zero" is well defined. Define $$ G : L\setminus\{x\} \to \mathbb{RP}^{n-1}_x, $$ where for each $y\in L$, $G(y)$ is the line in $\mathbb R^n$ passing though $x, y$. Again one can check that $G$ is smooth. Since $L$ has codimension $>1$, $\dim L < n-1=\dim \mathbb{RP}^{n-1}_x$. Thus the image is of measure zero.

Note that we use just a weak form of Sard theorem (about smooth maps $f : M \to N$ with $\dim M < \dim N$).