Definition: A function $f :[a, b] \Rightarrow\mathbb{R}$ is said to be piece-wise continuous if there exists a partition $a = t_{0} < t_{1} <...<t_m =b $, such that for each $k$, the restriction of $f$ to the open interval $(t_k,t_{k+1})$ is continuous, and extends to a continuous function in the closed interval $[t_k ,t_k + 1 ]$.
Definition: A function $g: [a, b] → R$ is called a step function if there exists a partition $(t_0, t_1, t_2,..., t_m)$ of $[a, b]$, and numbers $(c_0, c_1, c_2, ..., c_m)$, such that $g(x) = c$ for $t_j< x <t_{j+1}$, $j = 0,1,2,..., m - 1.$ In other words g is constant on each open interval $(t_{j} ,t_{j+1})$.
My question:
Considering and comparing of two these definitons can we say that step functions are piecewise continious on the closed intervals?
P.S.
as mentioned on the coment by @copper.hat "The only question here is if the constant functions can be extended to a continuous function on the closed interval". But in my opinion it is not so diffucult. Just we can define limit values to the end points.
(I'm unable to add to the thread as I don't yet have rep 50 :P)
An implicit part of your definition of piece-wise continuous is that the subfunctions, $f_k$, you have for each interval $[t_k, t_{k+1}]$ must agree on endpoints i.e.
$$f_{k}(t_{k+1}) = f_{k+1}(t_{k+1})$$
So consider what happens to these endpoints on a generic step function