Let f be a function defined on R by f(x)=[x]+√(x-[x]), then

$f(x)=n+\sqrt{x-n},\,n≤x<n+1$

If $x_0=K∈I$, then $\underset{x→K+}{\lim}f(x)=\underset{x→K}{\lim}K+\sqrt{x-K}=K$

and $\underset{x→K-}{\lim}f(x)=\underset{x→K-}{\lim}(K-1)+\sqrt{x-(K-1)}$

$=\underset{x→K}{\lim}(K-1)+\sqrt{x-(K-1)}=K-1+1=K$

Hence f is continuous at every $x_0=K∈I$. If $x_0∈R~ I$, then [x] is continuous at x₀ which is
turn gives that f is continuous at every $x_0∈R~ I$