Why is the upper sum of a refinement less than the upper sum of the original partition.

Question

In the lecture notes for my real analysis module, it is proven that: If we have a partition $P$ and $P'$ is a refinement of $P$ then $U(f,P') \le U(f,P)$. But why is this the case? My lecturer used the following proof:

I am unsure why $$\sum_{i=1,i \neq j}^{n}\sup_{x\in[x_{i-1},x_i]}f(x)(x_i-x_{i-1}) + \sup_{[x_{j-1,x_j}]}f(x)(x_j-x_{j-1}) \ge\sum_{i=1,i \neq j}^{n}\sup_{x\in[x_{i-1},x_i]}f(x)(x_i-x_{i-1})+ \sup_{[x_{j-1},{\gamma}]}f(x)(\gamma-x_{j-1})+ \sup_{[{\gamma},x_j]}f(x)(x_j-\gamma)$$ Can anyone explain to me why this line is true?

Answer 1

It boils down to $$\sup_{x\in A}f(x)\geq \sup_{x\in B}f(x)$$ whenever $B\subseteq A$. Which is basically obvious: $A$ has additional possible inputs when compared to $B$, so we can get $f(x)$ to be any size that it could be when $x\in B$, but we could also potentially get it to be larger than that. So we have that $X:=\sup_{[x_{j-1},x_j]}f(x)$ is greater than or equal to both $Y:=\sup_{[x_{j-1},\gamma]}f(x)$ and $Z:=\sup_{[\gamma,x_j]}f(x)$.

And then we have

$$\begin{align}X(x_j-x_{j-1})&=X(x_j-\gamma+\gamma-x_{j-1})\\ &=X(x_j-\gamma)+X(\gamma-x_{j-1})\\ &\geq Y(x_j-\gamma)+Z(\gamma-x_{j-1}) \end{align}$$

Answer 2

Suppose $\gamma$ was that point in $[x_{j-1},x_j]$ such that $\sup_{[x_{j-1},x_j]}f(x) = f(\gamma)$. Then $$\sup_{[x_{j-1},x_j]}f(x)(x_j - x_{j-1}) = f(\gamma)(x_j - x_{j-1}) = f(\gamma)(\gamma - x_{j-1}) + f(\gamma)(x_j - \gamma).$$ Otherwise (WLOG) $\sup_{[x_{j-1},x_j]}f(x)$ occurs at a point $x^* \in [x_{j-1},\gamma]$. Then \begin{align*}\sup_{[x_{j-1},x_j]}f(x)(x_j - x_{j-1}) = f(x^*)(x_j - x_{j-1}) &= f(x^*)(\gamma - x_{j-1}) + f(x^*)(x_j - \gamma) \\&\geq \sup_{[x_{j-1},\gamma]}f(x)(\gamma - x_{j-1)} + \sup_{[\gamma, x_j]}f(x)(x_{j} - \gamma).\end{align*}