# Strict inequality in proof of First Mean Value Theorem for Integrals

Let $M=\sup|f(x)|$ and $m=\inf|f(x)|$ $x \in [a,b]$

The first mean value theoremn for Riemann integrals says:

If $f$ continuous (and in this case we will asume non-constant, constant is trivial) then there is $\xi \in (a,b)$ (open interval) such that $$\int_a^bf(x)dx=f(\xi)(b-a)$$

The proof and the fact that it's an open inverval follows from this inequality: $$m(b-a)<\int_a^bf(x)dx<M(b-a)$$ which is strict (not $\leq$) because, let's say it can be equal:

$\int_a^bf(x)dx=M(b-a)=M\int_a^b dx \implies \int_a^b(M-f(x))dx=0 \implies M-f(x)=0 \implies M=f(x)$ so the equality is only true for constant functions. So it must be a strict inequality.

Now, first mean value theorem for Riemann-Stieltjes integrals says:

If $f$ is continuous and $\alpha$ increasing then there is $\xi \in [a,b]$ (closed interval) such that $$\int_a^bf(x)d\alpha=f(\xi)(\alpha(b)-\alpha(a))$$

In this case the inequality used in the proof is $$m(\alpha(b)-\alpha(a))\leq\int_a^bf(x)d\alpha\leq M(\alpha(b)-\alpha(a))$$

I'm not sure why in this case we can include the "equal to" case, I think it's because $\alpha$ can be super weird and constant in some places and discontinuous... So our teacher asked us, why is the inequality strict when $\alpha$ is continuous?

He gave us a hint but it doesn't use continuity of $\alpha$:

Let $P$ be a partition of $[a,b]$ and suppose that in $[x_i,x_{i+1}]$ $f(x)<M$ (which I think is possible if $f$ is not constant) and in the other intervals $f(x)\leq M$ so, $$\sum_{k=1}^n M_k \Delta x_k \leq \mu(x_{i+1}-x_i)+M\sum_{others}\Delta x_k < M(b-a)$$ with $\mu<M$

My question is: why is the inequality strict when $\alpha$ is continuous and not increasing.

I mean, it may be very easy but I'm so confused...this hint is not the full solution as I thought, he says...