Suz and Mike had a question

Problem Statement: If f(x) is bounded with finitely many discontinuities on [a,b] then f is Riemann Integrable on [a,b].

Proof: Let N be the number of discontinuities of f on [a,b] and let M\in\mathbb{R} be such that \left|f(x)\right|<M for every x\in [a,b]. We know such an M exists since f is bounded. Since f is bounded we will be using the sup-inf definition of Riemann Integrable.

Denote a discontinuity by d_j where d_1<d_2<\dots<d_N.  Let d=min\{\dfrac{\left|d_j-d_k\right|}{3}: j,k\in[1,N] and j\neq k\}. Let I_1=[a,d_1-d], I_2=[d_1+d,d_2-d],\dots I_{N+1}=[d_N,b]. By construction there are no discontinuities in any I_j and so f is Riemann Integrable one each I_j.

Let \varepsilon>0. Then for each I_j there exists a \delta_j>0 such that for any partition \pi of I_j with \|\pi\|<\delta_j it follows that \left|\sum\limits_{k=1}^{n}(sup-inf)\delta x_k\right|<\dfrac{\varepsilon}{2(N+1)}.

Let \delta=min\{\delta_1,\dots\delta_{N+1},d,\dfrac{\varepsilon}{4NM}\} and let \pi be a partition of [a,b] with \|\pi\|<\delta. Consider the sum below, we may separate the sum into parts, intervals which contain a discontinuity and those which do not.

\left|\sum\limits_{k=1}^{n}(sup-inf)\delta x_k\right|=\left|\sum\limits_{k=1}^{n_1}(sup-inf)\delta x_k+\sum\limits_{k=n_1}^{n_1+1}(sup-inf)\delta x_k+\dots+\sum\limits_{k={n_N+1}}^{n}(sup-inf)\delta x_k\right|


Note that there are N+1 I_j intervals and so there are N+1 terms less than \dfrac{\varepsilon}{2(N+1)}. Also note that we have bounded each of the intervals that contain a discontinuity by 2M\delta, this is since the largest possible difference between the sup and inf is 2M. Since there are N discontinuities there are N such terms in our inequality. Now we will use the fact that \delta<\dfrac{\varepsilon}{4NM} in order to simplify our inequality further.



Thus, we have shown that f is Riemann Integrable on [a,b].


This entry was posted in Analysis, Math, Riemann Integrable. Bookmark the permalink.

2 Responses to Suz and Mike had a question

  1. j2kun says:

    Thank goodness f(x) is founded, or this proof wouldn’t have anything to stand on! 😉

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