Let $f$ an function defined in $[a,b]$ bounded and continuous in all point of $[a,b]$ except in $x_0 \in [a,b]$ Prove f is integrated in $[a,b]$

I make this: Let $P=({t_0,.....,t_n})$ a partition of $[a,b]$. For $1\leq i \leq n$

Let $m_i=inf({f(x):t_{i-1} \leq x \leq t_i})$

$M_i=sup({f(x):t_{i-1} \leq x \leq t_i})$


$L(f,P)=\sum m_i(t_i - t_{i-1})$

I'm stuck, can someone help me?

  • $\begingroup$ Are trying to show that $f$ is (Riemann) integrable over $[a,b]$? If so, have you already done it for continuous functions? If yes, invoke linearity of the integral (which you have by definition). $\endgroup$ – Mathematician 42 May 17 '17 at 13:29
  • $\begingroup$ Yes I trying to prove riemman integrable! $\endgroup$ – Bvss12 May 17 '17 at 13:41
  • $\begingroup$ This can help. $\endgroup$ – Juniven May 17 '17 at 13:46

Let's first prove the following easy lemma:

Let $f$ be a real continuous function defined on some closed an finite interval $[a,b]$, then $f$ is Riemann integrable.

Idea of Proof. We have the show that the lower Riemann sum equals the upper Riemann sum. The lower Riemann sum is defined as follows: If $P=\left\{a=p_1, p_2, \dots ,p_{n-1},p_n=b\mid p_i<p_{i+1}\right\}$ is a partition of $[a,b]$ into finitely many segments, we define $L(f,P):=\sum_{i=1}^n \min(f_{\mid[p_i,p_{i+1}]})(p_{i+1}-p_i)$. Notice that $\min(f_{\mid[p_i,p_{i+1}]})$ is well-defined by the extreme value theorem. The lower Riemann sum is $L(f)=\inf_{P}L(f,P)$ where the infimum is taken over all partitions of $[a,b]$. Similarly one defines the upper Riemann sum $U(f)$.

Now one can order partitions. That is, we say that $P$ is finer than $P'$ if $P'\subset P$. Show that $P'\subset P\Rightarrow L(f,P)\geq L(f,P')$. For each $n$ one can define the partition $P_n=\left\{p_1, \dots ,p_n\right\}$ by $p_i=\frac{i(b-a)}{n}$. Show that $\lim_{n\rightarrow \infty}L(f,P_n)=L(f)$. Similarly results hold for upper sums.

Notice that when $n\rightarrow \infty$, $p_i\rightarrow p_{i+1}$ (Both are dependent on $n$ so you have to make this hard). Hence for any $x\in [p_i,p_{i+1}]$, $f(x)\rightarrow f(p_{i+1})$ as $n\rightarrow \infty$. In particular this holds for $x_{\min}$ and $x_{\max}$. It follows that $L(f)=U(f)$. $\square$

So how does this help? For any $\varepsilon>0$ (small enough) you can consider $f=f_1+f_2+f_3$ where $f_1=f_{\mid[a,x_0-\varepsilon]}$, $f_2=f_{\mid [x_0-\varepsilon,x_0+\varepsilon]}$ and $f_3=f_{\mid [x_0+\varepsilon,b]}$. Notice that $f_1$ and $f_3$ are continuous. Notice that $L(f)=L(f_1)+L(f_2)+L(f_3)$. Moreover, since $f_2$ is bounded (Why?) there exists an $\alpha$ (independent of $\varepsilon$) such that $L(f_2)<2\alpha \varepsilon$. Similarly, there is a $\beta$ such that $U(f)<2\beta\varepsilon$. By letting $\varepsilon\rightarrow 0$, we see that $L(f_2)\rightarrow 0\leftarrow U(f_2)$.

There is a lot going on here, writing everything down carefully can be quite annoying, but intuitively I'm only trying to point out that a single point can never change the value of an integral. It might be useful that you indicate to what extent you are familiar with Riemann sums.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.