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I need to prove that the integrability of a function doesn't change if we remove one point. Given is the function $f: [a; b] \rightarrow R$ which is integrable and $k \in [a,b]$. Let $m: [a,b] \rightarrow R$ be another function such that $f(x)=m(x)$ for all $x \in [a,b] \backslash \{ k\}$. We need to show that $m: [a,b] \rightarrow R$ is integrable with $\int_a^b \! f(x) \, \mathrm{d} x = \int_a^b \! m(x) \, \mathrm{d} x.$ I somehow have a conflict with myself since I think if we remove a point we would hurt the continuity of the function and thus loose its integrability. Yet again it is known that a function only needs to be continous on the intervall which one wants to integrate on. In other words one can split up the integral due to the linearity... So my attempt is even though the function $m(x)$ isn't defined at $k$ we can still calculate the area by splitting up the integral. $\int_a^k \! m(x) \, \mathrm{d} x + \int_k^b \! m(x) \, \mathrm{d} x = M(k) - M(a) + M(b) - M(k) = - M(a) + M(b) = M(b) - M(a)$
Taking into account that $m(x) = f(x)$ for all $x \in [a,b] \backslash \{ k\}$ and $\int_a^b \! f(x) \, \mathrm{d} x = F(b) - F(a) = M(b) - M(a)$ we should be done at this point. Am I on the right track?

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Using Riemann sums, show that a function that takes only one non-zero value over $[a,b]$ has integral $0$ over $[a,b]$. Your result will follow from this using linearity of the definite integral. –  David Mitra Dec 20 '12 at 19:07
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Note that, the function $m$ is defined at $k$ (as it should be if one is to argue that it is integrable over $[a,b]$). –  David Mitra Dec 20 '12 at 19:16
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Work from the definition. How do you modify a partition that is 'good' for $f$ into one that is 'good' for $m$. And vice versa? –  copper.hat Dec 20 '12 at 19:19
    
What I basically need to show now is that $\int_k^k \! f(x) \, \mathrm{d} x = 0$ and thus can be left out or am I mistaken? –  Just a Student Dec 20 '12 at 19:20

2 Answers 2

up vote 2 down vote accepted

The logical fallacy here is that continuity is only sufficient, but not necessary for Riemann Integrability. Now, in your proof you used the Fundumental Theorem of Calculus, which does require continuity and is thus invalid. The correct proof would involve the definition of the Riemann Integral.

Proof: As David Mitra pointed out, it suffices to prove the following:

Let $f:[a,b]\to \mathbb{R}$, $f(x)=0$ for $x\neq k$. Then $f$ is integrable in $[a,b]$ and $$\int_a^bf=0$$

Proof: Suppose WLOG $f(k)>0$, let $\epsilon>0$ and choose $n\in \mathbb{N}$ so that $$n>\frac{f(k)(b-a)}{\epsilon}$$ Let \begin{equation}\mathcal{P}=\left\{ a=x_0<x_1<...<a+\frac{i(b-a)}{n}<...<x_n=b \right\}\end{equation} be a partition of $[a,b]$. Then, $k\in (x_{r-1},x_r]$ for some $0< r< n$. Thus, \begin{equation} M_i(f)=\sup_{x\in [x_{{i-1}},x_i]}f(x)=\begin{cases} 0&\mbox{if, }i\neq r\\ f(k)&\mbox{if, }i=r\end{cases} \end{equation} and \begin{equation} m_i(f)=\inf_{x\in [x_{{i-1}},x_i]}f(x)=0 \end{equation} Therefore, \begin{equation} U_{f,\mathcal{P}}=\sum\limits_{i=1}^{n}{M_i(f)\left( x_i-x_{i-1} \right)}=f(k)(x_r-x_{r-1})=f(k)\frac{b-a}{n}<\epsilon \end{equation} while \begin{equation} L_{f,\mathcal{P}}=\sum\limits_{i=1}^{n}{m_i(f)\left( x_i-x_{i-1} \right)}=0 \end{equation} Therefore, $$U_{f,\mathcal{P}}-L_{f,\mathcal{P}}<\epsilon$$ and so $f$ is integrable by the Riemann criterion. Because $L_{f,\mathcal{P}}=0$, $$\int_a^bf=0$$

Note: If we are talking about Lebesgue Integration, this result is very simple since a finite set is countable and thus of measure $0$

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We didn't introduce Lebesgue Integration so I doubt that I need to prove it that way. However if I use the definition of the Riemann Integral I can only subdivide the sums a finite times. It will be an approximation but no real equalness. I am sorry but I have problems with the imagination of this certain idea... –  Just a Student Dec 20 '12 at 19:09

You could use the definition of a Darboux Integral to prove that $m$ is integrable.

Definitions

  1. The upper Darboux sum $U(f,P)$ of $f$ with respect to a partition $P$ is the sum $$U(f,P)=∑_{k=1}^nM(f,[t_{k-1},t_k])(t_k-t_{k-1})$$

  2. The lower Darboux sum $L(f,P)$ of $f$ with respect to a partition $P$ is the sum $$L(f,P)=∑_{k=1}^nm(f,[t_{k-1},t_k])(t_k-t_{k-1})$$

Theorem 3. A bounded function $f$ on$[a,b]$ is (darboux) integrable if and only if for each $\epsilon>0$ there exist a partion $P$ such that $$U(f,P)-L(f,P)<\epsilon$$

Let $B$ be a bound for $|f|$ and $|g|$. $B>0$, because it can't be that both function are the $0$ function. If $\epsilon>0$ there exist a partition $P$ such that

  1. $k$ belongs to one (and only one) interval $I$ in this partition
  2. $U(f,P)-L(f,P)<\frac{\epsilon}{3}$ (by theorem 3)
  3. The maximum length of interval is smaller then a positive number $δ$.

Consider now $U(m,P)-U(f,P)$:

$$|U(m,p)-U(f,P)|<M(f,I)⋅δ<[B-(-B)]⋅δ=2B⋅δ$$

We can choose this $δ$ smaller (as small as we want) and $(1)$ and $(2)$ will still hold. With the same argument we can make the difference between the lower bounds $|L(m,p)-L(f,P)|$ as small as we want. Note that:

$$U(m,p)-L(m,P)≤|U(f,p)-L(f,P)|+|L(m,p)-L(f,P)|+|U(m,p)-U(f,P)|$$

If we choose $δ=\frac{\epsilon}{6B}$, we get that $U(m,p)-L(m,P)<\epsilon$. Which proves that $m$ is integrable.

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