The following only proves that m(x) is continuous on [a, b], and the proof of M(x) is similar.

Any x0 belongs to [a, b]:

Case 1. f(x0) = M(x)，

Given e > 0, according to continuity, t > exists; 0, so | f (x)-f (x0) | < e. Then:

1.M(x)>f(x0) - e = M(x0) - e。

2. if x

So | m (x)-m (x0) | < e. So continuous.

Scene 2. f (x0)

According to the continuity, there is t 1 >: 0, so | f (x)-f (x0) | < M(x)-f(x0). In this neighborhood, M(x)=M(x0) is constant, so it is continuous.