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How can I show this integral is well defined?

Let f be an increasing function on [0,1] and $F(x)= \int _0^x f(t) dt$.

My attempt: f is an increasing function so it may only have a countable set of jump discontinuities:

$ m(x_0)=\inf f(x_0)\ \&\ M(x_0)=\sup f(x_0)\\ m(x_0)=M(x_0)$

Now I need to show it is bounded to conclude it has Riemann integral but I don't know how to show it.

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up vote 1 down vote accepted

An increasing function on a closed interval is of bounded variation. A function with bounded variation is bounded. If these concepts are unfamiliar to you, then I suggest taking a look at this reference.

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Since the function is increasing, one has that, given any partition $$P=\{t_0,\dots,t_n\}$$

$$M_i=\sup_{[t_{i-1},t_i]}f(x)=f(t_i)$$ $$m_i=\inf_{[t_{i-1},t_i]}f(x)=f(t_{i-1})$$

Then one has that the upper and lower sums are

$$U(f,P)=\sum_{i=1}^n M_i(t_{i}-t_{i-1})=\sum_{i=1}^n f(t_i)(t_{i}-t_{i-1})$$ $$L(f,P)=\sum_{i=1}^n m_i(t_{i}-t_{i-1})=\sum_{i=1}^n f(t_{i-1})(t_{i}-t_{i-1})$$

We then have that $$U(f,P)-L(f,P)=\sum_{i=1}^n (f(t_i)-f(t_{i-1}))(t_{i}-t_{i-1})$$

Let $\epsilon >0$ be given and let $P$ now be such that $t_i-t_{i-1}<\delta$ for some $\delta$ we will determine shortly. Then

$$U(f,P)-L(f,P)<\sum_{i=1}^n (f(t_i)-f(t_{i-1}))\delta$$

$$U(f,P)-L(f,P)<\delta \sum_{i=1}^n (f(t_i)-f(t_{i-1}))$$

$$U(f,P)-L(f,P)<\delta (f(t_n)-f(t_{0}))$$

Thus it suffices to take $\delta <\dfrac {\epsilon}{f(t_n)-f(t_{0})}$ for any given $\epsilon$ and thus we have integrabililty.

On a side note, if $f$ is increasing over $[a,b]$ it is immediately bounded since $f(a)\leq f(x)\leq f(b)$ for every $x\in [a,b]$.

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