More practice with proving a function is Riemann Integrable

If b\in [a,c] and if f is Riemann Integrable on both [a,b] and [b,c], then f is Riemann Integrable on [a,c].

Proof: Since we know f is Riemann Integrable on [a,b] and [b,c] it follows that f is bounded on those intervals as well. Let M be the maximum of those bounds, then it follows that |f(x)|\leq M  for every x\in [a,c].

Let \varepsilon >0. Since we know f is Riemann Integrable on [a,b] and [b,c], and bounded on [a,c], it follows that there exists a \delta_1>0 and \delta_2>0 such that for \pi_1 a partition of [a,b] with ||\pi_1||<\delta_1 and \pi_2 a partition of [b,c] with ||\pi_2||<\delta_2 it follows that |\sum\limits_{k=1}^n (\sup f[x_{k-1},x_k]-\inf f[x_{k-1},x_k])(x_{k}-x_{k-1})|<\dfrac{\varepsilon}{3}  where x_k\in \pi_1 and |\sum\limits_{k=1}^n (\sup f[y_{k-1},y_k]-\inf f[y_{k-1},y_k])(y_{k}-y_{k-1})|<\dfrac{\varepsilon}{3} where y_k\in\pi_2.

Let \delta=min\{\delta_1,\delta_2,\dfrac{\varepsilon}{12M}\}. Let \pi=\{x_0,x_1,\dots,x_n\} be a partition of [a,c] such that ||\pi||<\delta. Then b is in at most two intervals of \pi. This would happen if b is an endpoint of an interval. Suppose b\in [x_{m-1},x_m] and b\in[x_m,x_{m+1}] for some 0<m<n. Consider the sum below. We wish to show that the following sum is less than \varepsilon.

|\sum\limits_{k=1}^n (\sup f[x_{k-1},x_k]-\inf f[x_{k-1},x_k])(x_{k}-x_{k-1})|

=|\sum\limits_{k=1}^{m-1} (\sup f[x_{k-1},x_k]-\inf f[x_{k-1},x_k])(x_{k}-x_{k-1}) +\sum\limits_{k=m}^{m+1} (\sup f[x_{k-1},x_k]-\inf f[x_{k-1},x_k])(x_{k}-x_{k-1}) +\sum\limits_{k=m+2}^{n} (\sup f[x_{k-1},x_k]-\inf f[x_{k-1},x_k])(x_{k}-x_{k-1})|

<|\dfrac{\varepsilon}{3}|+|\sum\limits_{k=m}^{m+1} (\sup f[x_{k-1},x_k]-\inf f[x_{k-1},x_k])(x_{k}-x_{k-1})| + |\dfrac{\varepsilon}{3}|

\leq \dfrac{\varepsilon}{3}+ |\sum\limits_{k=m}^{m+1} (2M)(\delta)| +\dfrac{\varepsilon}{3}

\leq\dfrac{\varepsilon}{3} +4M\dfrac{\varepsilon}{12M} + \dfrac{\varepsilon}{3}



Thus, f is Riemann Integrable on [a,c].


Reflection: This is a type of proof that shows up a lot on the quals. The trick is that we know we can make most of the sum small since we know it’s RI on a part of the interval. Since our function is bounded this let’s us pick a delta that works nicely. Then we can make everything “small” and so it falls out from there.

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4 Responses to More practice with proving a function is Riemann Integrable

  1. j2kun says:

    Here’s a different perspective, using a slightly different definition (with, in my opinion, better notation).

    Recall that the definition of Riemann integrability on an interval [m,n] is that for any \varepsilon >0 there is a \delta > 0 such that any tagged partition P with norm less than \delta satisfies the following inequality:

    \displaystyle | \sum\limits_{i = 1}^{n} \Delta x_i t_i - L| < \varepsilon

    for some fixed finite limit L, with P having n intervals of length \Delta x_i, and tags t_i. This was standard notation in Dr. Shapiro’s analysis class, as the actual endpoints of the intervals are irrelevant unless the partition and function are explicitly specified.

    So given Riemann integrability on both of the above intervals, we may assert the existence of two limits L_1, L_2, such that for any epsilons \varepsilon_1, \varepsilon_2 > 0, there exist corresponding \delta_1, \delta_2 > 0, where for any tagged partitions P_1 = (\Delta x_i, t_i), P_2 = (\Delta y_i, s_i) with norms less than \delta_1, \delta_2, respectively, the following inequalities hold:

    \displaystyle | \sum\limits_{i = 1}^{n} \Delta x_i t_i - L_1| < \varepsilon_1 \\ \displaystyle | \sum\limits_{i = 1}^{m} \Delta y_i s_i - L_2| < \varepsilon_2

    with n, m the number of intervals of their respective partitions. Without loss of generality we may let the norm of both P_1, P_2 be less than \min{(\delta_1, \delta_2)}, and the inequality above still holds. Notice then, that P1 \cup P_2 is a tagged partition of [a,c] with norm less than both of \delta_1, \delta_2. Then the result follows as an application of the triangle inequality:

    \displaystyle  | \sum\limits_{i = 1}^{n} \Delta x_i t_i + \sum\limits_{i = 1}^{m} \Delta y_i s_i  - (L_1 + L_2) | \\  = | \sum\limits_{i = 1}^{n} \Delta x_i t_i - L_1 + \sum\limits_{i = 1}^{m} \Delta y_i s_i - L_2 | \\  \le  | \sum\limits_{i = 1}^{n} \Delta x_i t_i - L_1 | + | \sum\limits_{i = 1}^{m} \Delta y_i s_i - L_2 | \\ < \varepsilon_1 + \varepsilon_2

    As \varepsilon_1, \varepsilon_2 are both arbitrary, given any \varepsilon > 0 we may let \varepsilon_1 = \varepsilon_2 = \frac{\varepsilon}{97}, then we see that the Riemann sum above is within \frac{2 \varepsilon}{97}  \le \varepsilon, and the Riemann sum over the whole interval is L_1 + L_2.

    • j2kun says:

      correction: replace t_i with f(t_i), the values of the tags under f. Otherwise the Riemann sum S(f,P) doesn’t make sense at all 🙂

      And of course, if we don’t need to recall the definitions and notation, that cuts the proof shorter by three paragraphs. I was just trying to be overly pedantic.

  2. Suzanne Lavertu says:

    When I attacked this problem, I took the same approach as Jeremy, which I thought was easier due to the notation. But then again, a function being Riemann Integrable and bounded should give us a hint to use the sup/inf definition of Riemann Integrable, which has proved to be very helpful in other problems.

    • j2kun says:

      Right, but as per my and Erin’s discussion, the problem statement never mentions boundedness. That comes as a result of Riemann integrability.

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