# 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 $[a,b]$

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

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

Then

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

I'm stuck, can someone help me?

• 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). – Mathematician 42 May 17 '17 at 13:29
• Yes I trying to prove riemman integrable! – Bvss12 May 17 '17 at 13:41
• This can help. – 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.