The Gauss orthogonal polynomial has simple roots inside the interval

Call an interior root sign-changing if $p_m$ has opposite nonzero signs on the two adjacent intervals containing no root. Let $R$ be the set of such roots; it is finite by [F3]. Suppose, for contradiction, that $R$ has $k<m$ elements, and put $q(X)={\prod }_{s\in R}(X-s)$, with the empty product equal to $1$. Between consecutive roots of $p_m$, the product $h=p_{m}q$ is continuous and nonzero, so its image is an interval avoiding $0$ and its sign is constant by [F7]–[F8]. Crossing a root in $R$ flips both factors; crossing any other root of $p_m$ flips neither product sign. Thus $h$ has one weak sign on $[a,b]$.

Since $\mathrm{deg}q=k<m$, orthogonality [F2] gives $\int hd\mu =0$.

The polynomial $p_m$ has at most $m$ distinct roots by [F3]. The support condition [F1] therefore supplies a support point $y$ with $p_m(y)\mathrm{\ne }0$; as every root of $q$ is a root of $p_m$, also $h(y)\mathrm{\ne }0$. Continuity of $h$ gives a relative open neighborhood $U$ of $y$ on which $\mathrm{\mid }h\mathrm{\mid }\ge c$ for some $c>0$ and the sign of $h$ is unchanged. This neighborhood is Borel and $\mu (U)>0$ by the definition of support.

If $h\ge 0$ on $[a,b]$, [F4] gives $\int hd\mu \ge c\mu (U)>0$. If $h\le 0$, apply [F4] to $-h$ and use [F5] to obtain $\int hd\mu <0$. Both contradict step 2.1, so $k\ge m$.

By [F3], $p_m$ has at most $m$ distinct roots, and step 3.1 gives at least $m$, so $k=m$. Write them as $x_1,\dots ,x_m\in (a,b)$ and let $s(X)={\prod }_{i=1}^m(X-x_i)$. Both $s$ and $p_m$ are monic of degree $m$ by [F2, F6]. Their degree-$m$ coefficients cancel, so their difference is zero or has degree at most $m-1$ by [F9]. It vanishes at all $m$ distinct $x_i$; if nonzero, [F3] is contradicted. Thus $p_m=s$, so each root occurs as exactly one linear factor and is simple.