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Let $P$ be a partition of $[a, b]$ that is a refinement of the partition $P'$. For a real-valued function $f$ on $[a, b]$, show that $V(f, P') \leq V(f, P)$.

Proof: Let $P'$= {$a$ = $x_{0}$ < $x_{1}$ < ... < $x_{n-1}$ < $x_{n}$ = $b$} and $P$= {$a$ = $x_{0}$ < $x_{1}$ < ... < $x_{k-1}$ < $x_{k}$ < $x_{k+1}$ < ... < $x_{n-1}$ < $x_{n}$ = $b$}.

Then, $V(f, P')$ = $\sum^n_{i=1} |f(x_i)-f(x_{i-1})| = |f(x_n)-f(x_{n-1})| + ... + |f(x_2)-f(x_1)|$ and $V(f, P)$ = $\sum^n_{i=1} |f(x_i)-f(x_{i-1})| = |f(x_n)-f(x_{n-1})| + ... + |f(x_{k+1})-f(x_{k})|+ |f(x_{k})-f(x_{k-1})|+... + |f(x_2)-f(x_1)|$

Is this not as simple as the Triangle Inequality?

$|f(x_n)-f(x_{n-1})| + ... + |f(x_2)-f(x_1)| \leq |f(x_n)-f(x_{n-1})| + ... + |f(x_{k+1})-f(x_{k})|+ |f(x_{k})-f(x_{k-1})|+... + |f(x_2)-f(x_1)|$

Then, $V(f, P') \leq V(f, P)$?

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    Yes, that's all there is to it.2012-12-10

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