I know that any bounded function with a finite number of discontinuities is Riemann integrable over some interval. Is vice versa i.e.,  If a bounded function is Riemann integrable, then it has a finite number of discontinuities?  Thanks.
123107
Give example of a function $f:  [0,1] \to \mathbb R $ which is integrable ( Lebesgue or Riemann , if possible , both) but whose set of discontinuity points is  an uncountable set and dense in $[0,1]$ ?
115455
Is is true that if a function is Riemann integrable, then it is Lebesgue integrable with the same value? If it's true, how to prove it? If it's false, what is a counterexample?
39807
We know that the points of discontinuity of a monotone function on an interval $[a,b]$ are countable. Using this can we prov that:    Any bijection $f: \mathbb{R} \to [0,\infty)$ has infinitely many points of discontinuity.   If yes, how or otherwise how to prove the above result?
116903
Intuitively why must the primitive $F$ of a function $f$ that is Riemann integrable in $[a,b]$, be continuous in $[a,b]$?  Thanks
83306
Hello all mathematicians!! Again, I am struggling with solving the exercises in Lebesgue Integral for preparing the quiz. At this moment, I and my friend are handling this problem, but both of us agreed this problem is a bit tricky. The problem is following.   Let ($X,\mathcal{A},\mu$) be a measure space and suppose $\mu$ is $\sigma$-finite. Suppose $f$ is integrable. Prove that given $\epsilon$ there exist $\delta$ such that   $$ \int_A |f(x)|\mu(dx) \;\;&lt;\;\;\epsilon $$  whenever $\mu(A) &lt; \delta$.  Could anybody give some good idea for us? Think you very much for your suggestion in advance.
43637
Say I have a function $f(x)$ on some interval $[a,b]$.  Say it is integrable such that $\displaystyle\int f(x)~dx $ is defined.  Is $\displaystyle\int f(x)~dx $ necessarily continuous?  If I were to know that the integral is integrable itself such that $\displaystyle\int \int f(x)~dx $ is defined. Would that change anything?   If so why?  Thank you
49661
Prove that the bounded function $f$ defined by $f(x)=0$ if $x$ is irrational and $f(x)=1$ if $x$ is rational is not Riemann integrable on $[0,1]$.  I was given the hint to use the inverse definition of Riemann integrable and consider the cases of the partition being all rationals, and all irrationals between $[0,1]$, but I'm not too sure how to go about it.
120320
Let $f(x)$ be Riemann integrable on $[a,b]$ and satisfy $\inf\{\left|f(x)\right|:a\le x \le b\}=d&gt;0$.  How can we prove that $g(x)=\frac{1}{f(x)}$ is Riemann integrable?  I know from the statement that $f(x)\neq 0$ since its $\inf$ is $d&gt;0$, so $g(x)$ is bounded on $[a,b]$ since $f(x)$ is bounded and there is no division by $0$.    Is there a relationship like $M(f)=m(g)$ and $m(f)=M(g)$, since $g(x)$ is the reciprocal?
14373
$$\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$$  when f is continuous on $[0,1]$  I know it can be proved using bounded convergence theorem but,  I wanna know proof using only basic properties of riemann integral and fundamental theorem of calculus and MVT for integrals ...   Thank you.
19019
For an arbitrary interval $I$, how can we find a positive on $I$ integrable function? And how does one construct a measurable but not integrable function. If not all measurable functions are integrable, what is the best way to determine if a function is L-Integrable?   thx in advance
52979
Definitions  Suppose $P\colon a=x_0&lt;x_1&lt;\dotsb&lt;x_n=b$ is a partition of $[a,b]$. Let $\Delta x_k=x_k-x_{k-1}$ and $\lVert P\rVert$ denotes $\max_{0&lt;k\le n}\Delta x_k$.  The Cauchy integral of a function $f$ on closed interval $[a,b]$ equals to $I$ if and only if for each $\epsilon&gt;0$, there's some $\delta&gt;0$, for each partition $P$ of $[a,b]$ such that $\lVert P\rVert&lt;\delta$, we have $\left\lvert\sum_{k=1}^nf(x_k)\Delta x_k-I\right\rvert&lt;\epsilon$.  Problem  If $f$ is bounded on $[a,b]$ whose Cauchy integral equals to $I$, then $f$ is Riemann-integrable and $\int_a^bf=I$.  Background  It's an exercise from our calculus(analysis) problemset book, and there's a hint: consider the partitions whose $x_k-x_{k-1}$ is a constant for different $k$'s, and try to estimate the Riemann sum for each of these partitions through the Cauchy integral.  I have no idea about such estimation. After drawing some pictures, I discouraged. I googled on the Internet and found an . I realized that it's a quite different approach and with some advanced techniques (such as the analysis of a positive measure set -- discontinuities). I hope there will be some simpler approachers, just as the hint says. I need a more detailed hint, or a solution. Can anybody help me? Thanks!
109804
Give me an example of two Riemann-integrable functions $f,g:[0,1]\to[0,1]$ such that $g\circ f$ isn't integrable!  I already know the following example: $$f(x)=\begin{cases} 0,     &amp;     \text{if $x$ is irrational} \\ 1,     &amp;     \text{if $x=0$}\\ \frac1q, &amp;     \text{if $x$ is rational and $x=\frac pq$ such that $q\in\Bbb N$ and $(p,q)=1$}  \\ \end{cases}$$ $$g(x) = \begin{cases} 1,  &amp; \text{if $x$ is of the form $\frac 1q$such that $q\in \Bbb N$} \\ 0, &amp; \text{otherwise}  \\ \end{cases}$$ now observe that $g\circ f$ is a famous example of non-integrable function!
80664
Background  I am trying to prove the following theorem.     Let $f:[a,b]\to\mathbb{R}$ be a bounded function. If $c\in(a,b)$ then show that $f:[a,b]\to\mathbb{R}$ is Riemann Integrable on $[a,b]$ if $f$ is Riemann Integrable on both $[a,c]$ and $[c,b]$.   Notation  We use the following notations to simplify our discussion.   The collection of all partitions on $[a,b]$ for the function $f$ is denoted by $\mathcal{P}(f,[a,b])$. The upper sum for partition $P\in\mathcal{P}(f,[a,b])$ is denoted by $U(f,P,[a,b])$. The lower sum for partition $P\in\mathcal{P}(f,[a,b])$ is denoted by $L(f,P,[a,b])$.   Sketch of the Proof of the Theorem  Observe that, $$(P\in \mathcal{P}(f,[a,c]))\land(Q\in\mathcal{P}(f,[c,b]))\implies P\cup Q\in \mathcal{P}(f,[a,b])\tag{1}$$Furthermore, $$U(f,P\cup Q,[a,b])=U(f,P,[a,c])+U(f,Q,[c,b])\tag{2}$$$$L(f,P\cup Q,[a,b])=L(f,P,[a,c])+L(f,Q,[c,b])\tag{3}$$If we define, $$\mathcal{P}^\ast(f,[a,b]):=\{P\cup Q: (P\in \mathcal{P}(f,[a,c]))\land(Q\in\mathcal{P}(f,[c,b]))\land (P\cap Q=\{c\}) \}$$Then $\mathcal{P}^\ast(f,[a,b])\subseteq \mathcal{P}(f,[a,b])$. Now, observe that, \begin{align}\int_{a}^{\bar{b}}f&amp;=\inf\{U(f,R,[a,b]):R\in\mathcal{P}(f,[a,b])\}\\&amp;\le \inf\{U(f,R,[a,b]):R\in\mathcal{P}^\ast(f,[a,b])\}\\&amp;=\inf\{U(f,P(R),[a,c])+U(f,Q(R),[c,b]):(P(R),Q(R))\in\mathcal{P}(f,[a,c])\times \mathcal{P}(f,[c,b])\}\\&amp;=\inf\{U(f,P,[a,c])+U(f,Q,[c,b]):(P,Q)\in\mathcal{P}(f,[a,c])\times\mathcal{P}(f,[c,b]) \}\\&amp;=\inf\{U(f,P,[a,c]):P\in\mathcal{P}(f,[a,c])\}+\inf\{U(f,Q,[c,b]):Q\in\mathcal{P}(f,[c,b])\}\\&amp;=\int_{a}^c f+\int_{c}^b f\end{align}Similarly, \begin{align}\int_{\bar{a}}^{b}f&amp;=\sup\{L(f,R,[a,b]):R\in\mathcal{P}(f,[a,b])\}\\&amp;\ge \sup\{L(f,R,[a,b]):R\in\mathcal{P}^\ast(f,[a,b])\}\\&amp;=\sup\{L(f,P(R),[a,c])+L(f,Q(R),[c,b]):(P(R),Q(R))\in\mathcal{P}(f,[a,c])\times \mathcal{P}(f,[c,b])\}\\&amp;=\sup\{L(f,P,[a,c])+U(f,Q,[c,b]):(P,Q)\in\mathcal{P}(f,[a,c])\times\mathcal{P}(f,[c,b]) \}\\&amp;=\sup\{L(f,P,[a,c]):P\in\mathcal{P}(f,[a,c])\}+\sup\{L(f,Q,[c,b]):Q\in\mathcal{P}(f,[c,b])\}\\&amp;=\int_{a}^c f+\int_{c}^b f\end{align}Consequently we have, $$\int_{a}^c f+\int_{c}^b f\le \int_{\bar{a}}^{b}f\le \int_{a}^{\bar{b}}f\le  \int_{a}^c f+\int_{c}^bf$$which implies, $$\int_{a}^{b}f=  \int_{a}^c f+\int_{c}^bf$$  Question  Is my proof correct?
4575
I'm trying to prove  the following claim:     If $f$ is continuous  and $\displaystyle\int_a^\infty |f(x)|\;dx$ is finite then $\lim\limits_{ x \to \infty } f(x)=0$.   Here the counter example of all these sine functions won't do, since  their infinite integral does not converge absolutely.   Any hints for a proof?
115504
Problem: Suppose that $f$ is a bounded, real-valued function on $[a,b]$ such that $f^2\in R$ (i.e. it is Riemann-Integrable). Must it be the case that $f\in R$ ?      Thoughts: I think that this is not necessarily true, but I am having trouble refuting or even proving the above. Of course, the simplest way to prove that it is not necessarily true would be to give an example, but I am unable to think of one! I also have tried using $\phi(y)=\sqrt y$ and composing this with $f^2$ (to try show $f$ is continuous); however, the interval $[a,b]$ may contain negative numbers so I can't utilise $\phi$ in that case.  Question: Does there exist a function $f$ such that $f^2\in R$ but $f$ $\not\in R$ ? Or conversely, if $f^2\in R$ does this always imply $f$ $\in R$ ? (If so, could you provide a way of proving this).
81593
Let $f$ be a real, continuous function defined for all $0\leq x\leq 1$ such that $f(0)=1$, $f(1/2)=2$, and $f(1)=3$. Show that  $$\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$$  exists and compute the limit.  Attempt:   Since $f$ is real-valued continuous on $0\leq x\leq 1$ and the boundaries $f(0)=1$ and $f(1)=3$, $f$ is bounded on the interval  and the integral $\displaystyle\int_0^1 f(x^n)dx$ exists for any positive $n$. Thus, we can interchange the limit and the integral and compose the limit, \begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&amp;=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &amp;=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &amp;=\int_0^1 f(0)dx=1. \end{align*}  Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where $f(x^n)$ is dominated by a function $g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)$ and $f_n(x^n)$, on $0\leq x\leq 1$, converges pointwise to a function $f$ that takes value 1 for $x\neq 1$ and 3 for $x=1$.  My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified.
9845
Let $f$ be a real valued function on $[0,\infty)$ defined by  $$f(x)= \begin{cases} x^{{2}/{3}}\log x &amp; \text{ if } x&gt;0 \\  0 &amp; \text{ if } x= 0 \end{cases}$$  Then which of the following is true?  A. $f$ is discontinuous at $x=0$.  B. $f$ is continuous on $[0,\infty)$, but not uniformly continuous on $[0,\infty)$.  C. $f$ is uniformly continuous on $[0,\infty)$.  D. $f$ is not uniformly continuous on $[0,\infty)$, but uniformly continuous on $[1,\infty)$.  I could see that $f(x)=x^{\frac{2}{3}}\log x$ is continous at $x=0$   But for uniform continuity I thought boundedness of derivative would help   But I found $f'(x)=\frac{2}{3}x^{-\frac{1}{3}}\log x + x^{-\frac{1}{3}}$   This $f'(x)$ is not bounded so i can not say any thing about uniform contunuity.  Please help me to clear this.   Thank you :)  P.S : This was already asked before ( some other person but it was not having enough details so it got closed. I tried to edit that question but someone said I should not edit OP to add more details. I was afraid that may be against the rules of MSE. So, I thought i should ask this separately with my own ideas.