# Rigorous Proof?: Proving Cauchy Criterion of Integrals

I am trying to prove the following: (I only know about Riemann Integrals)

Let $f$ be a bounded function on an interval [a, b] and assume that $P_n$ is a sequence of partitions of [a, b] such that $\lim\limits_{n\to\infty} L(f, P_n) = \lim\limits_{n\to\infty} U(f, P_n) = s.$

Then, $f$ is integrable and $\int_a^bf dx = s.$

Proof:

We know that if $f$ is bounded on $[a,b]$, then $f$ is integrable on [a,b] if and only if for every $\varepsilon \gt0$ there is a partition $P$ of $[a,b]$ such that $U(f,P)−L(f,P)\lt\varepsilon.$

From our assumption, we know that there exists a partition $P_n$ s.t. $U(f,P_n)−L(f,P_n)=s-s=0\lt\varepsilon$ when $n\to\infty$

Therefore, $f$ is integrable.

Since $f$ is integrable, $L(f,P)=U(f,P)= \int_a^bf dx$,

$\int_a^bf dx=s$

Question (Comment):

I feel like what I have is very "weak" and unclear. I'm also unsure whether or not the last part even makes sense. Thank you.

Notation: U(f,P) indicates the Riemann Upper Sum and L(f,P), the lower sum.

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Is there any statement about the mesh size of $P_n$, or do you have to deduce it yourself? –  Arturo Magidin Nov 22 '11 at 22:15
Just because $\lim\limits_{n\to\infty} L(f, P_n) = \lim\limits_{n\to\infty} U(f, P_n) = s.$ doesn't mean that there is a specific $P_n$ such that $U(f,P_n)=L(f,P_n)=s$. In fact, usually this is not true, unless $f$ is a essentially a step function. –  Thomas Andrews Nov 22 '11 at 22:23
@Arturo No. Wouldn't the above equality of the limits mean that the mesh size = 0 though? –  MathMathCookie Nov 22 '11 at 22:24
@MathMathCookie: Either that, or that $f$ is constant. But my question was rather to know whether you have to prove it, or you are given it. –  Arturo Magidin Nov 22 '11 at 22:25
@MathMathCookie I think you need to modify the argument as follows: There is a partition $P_L$ that gets the lower sum within $\epsilon/2$ and $P_U$ which gets the upper sum within $\epsilon/2$. Then let $P$ be a common refinement of these two partitions. In that case $P$ will get you within $\epsilon/2$ of both the upper and lower sums and so they must be no more than $\epsilon$ apart. –  Bill Cook Nov 22 '11 at 22:34

For every $\varepsilon\gt0$, one wants to exhibit a partition $P$ such that $U(f,P)-L(f,P)\lt\varepsilon$.
Since $U(f,P_n)\to s$ when $n\to\infty$, there exists $n_U(\varepsilon)$ such that for every $n\geqslant n_U(\varepsilon)$, $|U(f,P_n)-s|\lt\frac12\varepsilon$. Since $L(f,P_n)\to s$ when $n\to\infty$, there exists $n_L(\varepsilon)$ such that for every $n\geqslant n_L(\varepsilon)$, $|L(f,P_n)-s|\lt\frac12\varepsilon$.
Choose $n^*(\varepsilon)=\max\{n_U(\varepsilon),n_L(\varepsilon)\}$ and $P=P_{n^*(\varepsilon)}$. Then $$U(f,P)-L(f,P)\leqslant |U(f,P_{n^*(\varepsilon)})-s|+|L(f,P_{n^*(\varepsilon)})-s|\lt \tfrac12\varepsilon+\tfrac12\varepsilon=\varepsilon.$$ This proves that $f$ is integrable.
Furthermore, $L(f,P_n)\leqslant \int\limits_a^b f\leqslant U(f,P_n)$ for every $n$, hence $$\sup\limits_n\ L(f,P_n)\leqslant \int\limits_a^b f\leqslant \inf\limits_n\ U(f,P_n).$$ The supremum on the LHS and the infimum on the RHS are both equal to $s$ hence $\int\limits_a^b f=s$.