If $\lim_{x\to c}f(x)$ limit exists at every point $c\in[a,b]$, then $f$ is integrable over $[a,b]$.

Question

Berkeley problems problem 1.5.4

Suppose $f$ is real valued function of one real variable such that $\lim_{x\rightarrow c}f(x)$ exists for all $c\in [a,b]$. Show that $f$ is Riemann integrable on $[a,b]$.

Consider a jump discontinuity $c$. There exists a neighborhood s.t. image of a given point in this neighborhood is close to $\lim_{x\rightarrow c}f(x)$. So while calculating $U(P, f)$ and $L(P, f)$, the actual value of $f(c)$ may be ignored. But how to justify this?

One way is to consider a partition $P$ which contains $c_1$ and $c_2$ situated close to $c$ and on either side of $c$.

Please give a hint. Please do not give solution. Thanks!

If $\lim_{x\to c} f(x)$ exists for all $c \in [a,b]$ it means that $f$ is continuous, so what's the problem? — Gabrielek, Jun 30 '20 at 15:52
@Gabrielek it requires ${\lim_{x\rightarrow c}f(x)=f(c)}$, which is not implied by the limit simply existing — Riemann'sPointyNose, Jun 30 '20 at 15:54
It doesn't imply $f$ is continuous, since it could have point discontinuities. — LostStatistician18, Jun 30 '20 at 15:54
Possible duplicate of https://math.stackexchange.com/questions/3119713/is-a-function-with-limits-riemann-integrable — Mark, Jun 30 '20 at 15:57

score 1 · Accepted Answer · answered Jun 30 '20 at 15:59

HINT:For each $c \in [a,b]$, the hypothesis can be used to show that there exits a $\delta_c >0$ such that $|f(x)-f(y)| < \epsilon$ if $x,y \in (c-\delta_c,c+\delta_c) \setminus\{c\}$. The intervals of the form $I_c = (c-\delta_c,c+\delta_c)$ form an open cover for $[a,b]$. Using compactness, finitely many of them say $(c_1-\delta_{c_1},c_1+\delta_{c_1}),\dots,(c_n-\delta_{c_n},c_n+\delta_{c_n})$ cover $[a,b]$. Now choose a partition with $x_0 = a,x_1 = c_1,\dots,x_{n-1} = c_n,x_n = b$.

If $\lim_{x\to c}f(x)$ limit exists at every point $c\in[a,b]$, then $f$ is integrable over $[a,b]$.

1 Answers1