I want to show that if $f$ is integrable on the interval $[0,1]$ then $\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n f\left(\frac{k}{n}\right)=\int_0^1f$. I am using the definition of integrability in the sense of Riemann-Stieltjes: $$\forall\epsilon&gt;0:\exists P:\forall Q\geq P:\left|S(Q_n,f,I)-\int_0^1f\right|&lt;\epsilon$$ Where $P$ and $Q$ are partitions of $[0,1]$. I thought about using the definition for the partitions: $$Q_n=\left(0,\frac1n,\frac2n,\cdots,\frac{n-1}n,1\right)$$ $$c_k=\frac kn$$ Then we have: $$\left|S(Q_n,f,I)-\int_0^1f\right|=\left|\frac1n\sum_{k=1}^nf\left(\frac kn\right)-\int_0^1f\right|$$ The problem is that I don't know how I can show that this tends to zero since a $P$ appears in the definition of integral and I don't know what to do with it.
123888
I need to prove that if $f$ is Riemann integrable $[0,1]$ then $$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum f\left(\frac{k}{n}\right)= \int\limits_{0}^{1} f(x)dx$$  My idea is to recognize the right side defining the limit as a Riemann sum and use the uniform continuity.  I am not sure what to do next  Thanks
30814
Suppose $f$ is a Riemann integrable function on $[0,1]$. Prove that $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$   This is what I am thinking: Fix $n$. Then by Jensen's Inequality we have $$0\leq\left(\int_0^1x^nf(x)dx\right)^2 \leq \left(\int_0^1x^{2n}dx\right)\left(\int_0^1f^2(x)dx\right)=\left(\frac{1}{2n+1}\right)\left(\int_0^1f^2(x)dx\right).$$Thus, if $n\to\infty$ then $$0\leq \lim_{n\to \infty}\left(\int_0^1x^nf(x)dx\right)^2 \leq 0$$ and hence we get what we want. How correct (or incorrect) is this?
123995
I need help proving that if f is Lebesgue integrable then $$\sum_{k \in \mathbb{Z}}{ 2^k*m(|f| &gt; 2^k)}$$  converges.   I proved the opposite direction but I don't know how to put any bounds on each term, I would appreciate hints/suggestions.
108075
The function $f(x) = \sin(x)/x$ is Riemann Integrable from $0$ to $\infty$, but it is not Lebesgue Integrable on that same interval.  (Note, it is not absolutely Riemann Integrable.)  Why is it we restrict our definition of Lebesgue Integrability to absolutely integrable?  Wouldn't it be better to extend our definition to include ALL cases where Riemann Integrability holds, and use the current definition as a corollary for when the improper integral is absolutely integrable?
39620
Source: Old comp./preliminary exam.  Let $f(x)$ be a Riemann integrable function on $[0,1]$. Prove that $$\lim_{m\to\infty}\int_{0}^{1}f(x)\cos(mx) \, \,dx=0$$  Thought  $(1)$ Because we don't know if $f(x)$ is differentiable, we can only use integration by part by setting $du=f(x)\, \,dx$ and $v=\cos(mx)$, the result is not quite helpful.  $(2)$ We don't know whether $f(x)\cos(mx)$ converges as $m\rightarrow \infty$ so those convergence theorems cannot be applied.  Side note: This exam assume no knowledge in measure theory and Lebesgue integral.
53732
Evaluate $$\lim_{n\to\infty}\frac {qn+1}{qn} \frac {qn+p+1}{qn+p} \dotsm \frac {qn+np+1}{qn+np}$$ where $p\in\mathbb{N}\setminus\{0,1\},q&gt;0.$   Any hints on how to approach this problem in the first place? The answer should be: $\left(\frac {p+q}q\right)^{1/p}$  I have just tried something and got pretty close to the result anyway:  Let $\lim_{n\to\infty} \frac {qn + 1}{qn}\frac {qn+p+1}{qn+p}\dotsm\frac {qn+np+1}{qn+np}=L$ taking the $\ln$ on both sides we get:  $$\lim_{n\to\infty} \sum_{k=0}^n\ln\left(1+\frac1{qn+kp}\right).$$  Here (I'm not sure if I'm wrong here) but if we use the remarkable limit on each of these $\ln$. $\lim_{x\to0}\frac{ln(1+x)}{x}=1$.  $$\lim_{n\to\infty}\sum_{k=0}^n\frac1n\frac1{q+ \frac knp}\xrightarrow[{n\rightarrow\infty}]{}\int_0^1\frac 1{q+xp}dx=\frac 1p\ln\frac {p+q}{q}=\ln L.$$  $\implies \boxed{L=\sqrt[p]{\frac{p+q}q}}.$
122687
Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a&gt;0$, show that:  $$ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$$  Tried using Riemann sum, but did not succeed.
72681
Let $I$ b a generalized rectangle in $\Bbb R^n$   Suppose the bounded function $f:I\to \Bbb R$ assumes the value $0$ except at a single point $x \in I$   Show that $f$ is integrable and $\int_{I} f=0$     I think that   Let $\epsilon &gt;0$   Let's choose a partition $$\vert \vert P\vert\vert ^n&lt; \frac{\epsilon}{2\vert f(x)\vert} $$  and x contained in exactly one generalized rectangle in $P$.   I need to calculate $L(f,P)$ and $U(f,P)$.   How to calculate $L$ and $U$? Please help me for the evaluation. Thank you.
51322
Could please give me a hint.  $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k^{2}+1}$$  It looks like an integral, but I failed to figure out the function.
101310
The following problem came up on a test, and I didn't get it correctly. Please help me understand the problem.  Problem: Define $f:[0,1] \to \mathbb{R}$ by $$     f(x)=  \begin{cases}     1,&amp; \text{if } x=\frac{1}{n}\\     0              &amp; \text{otherwise} \end{cases} $$ for some integer n.  Using the definition of Riemann integrability, prove that f is Riemann integrable on $[0,1]$ and evaluate the Riemann integral $\int_{0}^{1} f(x)dx$.  My Solution:  Let $\epsilon&gt;0$ and $P$ be the partition of $[0,1]$ such that $P=\{0,\frac{1}{n}- \frac{\epsilon}{3},\frac{1}{n}+ \frac{\epsilon}{3},1\}$.Then note that   $U(f,P)=M_0\Delta x_0 +M_1\Delta x_1+M_2\Delta x_2=0+(\frac{1}{n}+ \frac{\epsilon}{3})-(\frac{1}{n}+ \frac{\epsilon}{3})+0=\frac{2\epsilon}{3}$ and   $L(f,P)=0$.   Thus, for every $\epsilon&gt;0$ $$U(f,P)-L(f,P)=\frac{2\epsilon}{3}&lt;\epsilon$$.  This is the prove that I gave and the mistake that I made was that I fixed n. Some important things that I noticed about this problem were that if you graph the function $f(x)$ most of the points fall to the left or close to zero and a fewer fall to the right. Hence, this makes me believe that I need a partition such that I capture most of the points near zero so that I am only left with a finite number of points close to 1. However, I can't seem to find this partition.Any help would be appreciated. In particular, I would like to know how you come up with the partition since I always struggle to come up with one. Thank you!
21265
Suppose $f : [ 0, 1 ] → \mathbb{R}$ and $\lim\limits_{x \rightarrow c} f ( x )$ exists for all $c \in [ a , b ]$. Show that $f$ is Riemann integrable on $[ a , b ]$.  I can show this $f$ is bounded on $[a,b]$ since the limit exists at every point. But I'm not sure how to proceed. Any hints or help would be great.
67111
Is the function Riemann integrable? If yes, obtain the value of $\int_0^1f(x) \ dx$      $f(x) = \begin{cases} \frac {1}{n},  &amp; \frac{1}{n+1}&lt;x\le\frac{1}{n}\\ 0, &amp; x=0 \end{cases}$   My attempt  $f$ is bounded and monotonically increasing on $[0,1]$. Also, $f$ has infinite discontinuities but only one limit point. Therefore $f$ is Riemann integrable. Now, to calculate the integration  $\int_0^1f(x) \ dx=\int_{1/2}^{1}1 \ dx + \int_{1/3}^{1/2}\frac{1}{2} \ dx + \int_{1/4}^{1/3}\frac{1}{3} \ dx+...$  $=\sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1}$  How do I proceed from here? How do I calculate these summations? I know $\sum \frac{1}{n}$ is $\log 2$, but not the other two summations.
18432
Given that $f:[0,1]\to \Bbb{R}$ is a continuous function, I need to show that $$\lim_{n \to \infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=e^{\int_0^1f(x)dx}$$  Writing $$y=\lim_{n \to \infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)$$ and taking the logarithm of both sides and interchanging limit and logarithm(since $f$ is continuous), I get $$\log y = \lim_{n\to \infty}\sum_{k=1}^{n}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)$$ Now I know that the limit of the Riemann sum $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{b-a}{n}f\left(a+\frac{(b-a)}{n}k\right)=\int_a^bf(x)dx$. However if I want it to read $\int_0^1f(x)dx$ in the given problem, I would need the term inside the summation to read $\frac{1}{n}f\left(\frac{k}{n}\right)$. The way to do this it would seem is to take the series expansion of $\log(1+x)$ and ignoring the $x^2$ and higher order terms. Is it justifiable to do so?
107794
Prove for bounded $f$, if $f:[a,r] \to \mathbb{R}$ is Riemann-Integrable for $r \in [a,b)$ then $f:[a.b] \to \mathbb{R}$ is Riemann-Integrable and $\displaystyle \lim_{r \to b, r &lt; b} \int_a^r f= \int_a^bf$.  Alternative definition for Riemann-Integrable functions :: $f : [a,b] \to \mathbb{R}$ is Riemann-Integrable if $\forall \varepsilon &gt; 0, \exists$ step functions $f_\varepsilon, g_\varepsilon$ such that $\lvert f- f_\varepsilon \rvert \le g_\varepsilon$ and $Ig_\varepsilon \le \varepsilon$ where $Ig_\varepsilon = \sum_{j = 0}^{N}g_\varepsilon(x)(\Delta x_{j+1} - \Delta x_j)$.  $f$ is bounded so $\displaystyle \exists M \in \mathbb{R}^+$ s.t. $|f(x)| \le M, \forall x \in [a,b] $ so we know  $\displaystyle|\int_a^r f| \le \int_a^r|f| \le M(r-a) \le M(b-a) $. Now, for some index set $I$ where $r_0 \le r_1 \le \ldots\le r $ we have $\displaystyle \int_a^rf = \sum_{i \in I}\int_{a}^{r_{i}}f$  Do I need to show $f$ is Riemann-Integrable over $[a,b]$ before I can solve the limit? It seems like it but seems like the other way would give it to me.  EDIT:  Since $f(x) \le M, \forall x \in [a,b]$ and we have there exists $r \in [a,b]$ such that $b - \frac{1}{n+1} &lt; r &lt; b$ we have $\displaystyle \int_a^{b-\frac{1}{n+1}}|f| \le \int_a^r|f| \le M(b-a)$ Does this give me that $f$ is Riemann-Integrable over $[a,b]$?
86068
Let $f_n$ be a sequence of differentaible functions on $[0,1]$ to $\mathbb{R}$ converging uniformly to a function $f$ on $[0,1]$, Then    $f$ is differentiable and Riemann integrable there $f$ is uniformly continuous and R-integrable $f$ is continuous, need not be differentiable on $(0,1)$ and need not be R-integrable on $ [0,1]$, $f$ need not be continuous.   Well I think 2 is only correct statement  as every $f_n$ are differentiable on $[0,1]$ hence uniformly contnous, and as they converge uniformly to $f$ on $[0,1]$ so $f$ is uniformly continuous and every continuous function is rieman integrable..Am I correct?
106583
Let $(X,\mathcal{M},\mu)$ be a measure space and $f$ be a nonnegative measurable function on $X$. Let $1\le p&lt;\infty$. Show that, the function $|f|^p$ is integrable with respect to $\mu$ precisely when the function $$t\mapsto t^{p-1}\mu({x:|f(x)&gt;t})$$ is integrable on $[0,\infty)$ with respect to Lebesgue measure. In addition prove that    $$\int_X |f|^p=p\int^{\infty}_{0} t^{p-1}\mu({x: |f(x)&gt;t}) dt\,$$   I solved this problem for the case when space is sigma finite but don't know how to solve for general case.
17745
I found this problem on an old FB post and have just been humbled by it.  Let $f(x)$ be continuous on $[0,n]$ such that $n &gt; 0$ and $f(x) + f(n - x)$ does not vanish on $[0,n]$ then evaluate  $$ \int_0^n \frac{f(x)}{f(x) + f(n-x)}dx $$  Any clues/hints would be appreciated.