If f(x) and g(x) are Riemann integrable in [a,b], why $h(x)=\max\{f(x),g(x)\}$ is still  Riemann integrable in [a,b]?  Or maybe it is wrong?
31774
Let $f,g$ be Riemann integrable functions, prove that the function $ h(x) $ defined by $$ h\left( x \right) = \max \left\{ {f\left( x \right),g\left( x \right)} \right\} $$ is also Riemann integrable.
114627
Let $f:[a,b]\to  R$ be integrable and continuous on $(a,b)$. Prove that: $$\lim_{h\to 0} \int_a^b\frac{f(x+h)-f(x)}{h} \,\mathrm{d}x = f(b)-f(a).$$  What I've tried so far: make a change of variables, which didn't help. I also thought of using the intermediate value theorem inside the integral, but the function being differentiable is not specified. Any help, tips? Thanks in advance!
106246
Problem: Let $g:[0,1]\to\mathbb{R}$ be a continuos function such that $g(1)=0$. Show that $\sup\limits_{x\in[0,1]}|x^ng(x)|\to0$ as $n\to\infty$.  My approach: Let $f(x)=x^ng(x)$. Then, $f(x)$ is continuos on $[0,1]$. Using the extreme value theorem we can say, that $|f(x)|$ is bounded on $[0,1]$ and attains its bounds, i.e., $\exists x_0\in[0,1]$ such that $|f(x_0)|=\sup\limits_{x\in[0,1]}|f(x)|$.  Now, $|f(x_0)|=|x_0^ng(x_0)|$. If $x_0=0$, then we are done. So, let us assume $x_0\neq0$. By the Archimedean property of $\mathbb{R}$, $\exists$ a $y\in\mathbb{R}$ such that $y=x_0^n\leq1$, since $x_0\in[0,1]$.   Hence, $\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|x^ng(x)|=\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|f(x)|=\lim\limits_{n\to\infty}|f(x_0)|=\lim\limits_{n\to\infty}|yg(y^{\frac{1}{n}})|\leq\lim\limits_{n\to\infty}|g(y^{\frac{1}{n}})|=g(1)=0.$  I will be glad if anyone can tell me if I have done any mistake and if so please help me to correct it.  Thank you.
26140
Question:  If $f\in C^1[0,1]$, show that  $$\int_0^1 |f(x)|dx\le\max\left\{\int_0^1 |f&#39;(x)|\,dx,\;\bigg|\int_0^1 f(x)\,dx\bigg|\right\}.$$  I have tried to make connection between $|f|$ and $|f&#39;|$ by using  $$(tf(t))&#39;=f(t)+tf&#39;(t),$$  integrate the equation with $t$ on $[0,1]$, it gets  $$|f(1)|\leq \bigg|\int_0^1 f(t)\,dt\bigg|+\int_0^1 |f&#39;(t)|\,dt,$$  but it is not the desired inequality.
107972
Theorem says that if $f:[a,b]\longrightarrow \mathbb{R}$ is a continuous function, then it's integrable in Riemann's sense. Here's my proof:  Proof  Let $\mathfrak{P}=\{x_0,x_1,...x_n\}$ be a subdivision of $[a,b]$. Let denote $m_k=\inf_{x\in[x_{k-1},x_k]}f(x)$, $M_k=\sup_{x\in[x_{k-1},x_k]}f(x)$ and a diameter as $\max_{k=0,1...,n}(x_{k+1}-x_k)$. Then, the lower Darboux sum is $s(f,\mathfrak{P})=\sum_{k=1}^n m_k(x_k-x_{k-1})$ and upper sum is $S(f,\mathfrak{P})=\sum_{k=1}^n M_k(x_k-x_{k-1})$.  We know that $f\in\mathfrak{R}([a,b])\iff \forall_{\epsilon &gt;0}\exists_{\mathfrak{P}}:S(f,\mathfrak{P})-s(f,\mathfrak{P})&lt;\epsilon$  Every funtion continuous on closed interval is also uniformly continuous, so $$\forall_{\epsilon &gt;0}\exists_{\delta &gt;0}\forall_{x_1,x_2\in [a,b]}:|x_1-x_2|&lt;\delta \Rightarrow |f(x_1)-f(x_2)|&lt;\frac{\epsilon}{b-a}$$  This means that for every subdivision $\mathfrak{P}$ of a diameter smaller than $\delta$ we get $$S(f,\mathfrak{P})-s(f,\mathfrak{P})=\sum_{k=1}^n(M_k-m_k)(x_k-x_{k-1})\leq \frac{\epsilon}{b-a}(b-a)=\epsilon$$ which ends the proof.  Is it understandable and, what's more important, correct?
94143
Let the functions $f$ and $g$ be holomorphic in $U$ and continuous in $\overline{U}$. Show that $|f(z)| + |g(z)|$ attains its maximum on $\{|z| = 1\}$. Hint: consider the function   $h = e^{iα}f + e ^{iβ}g$ with suitably chosen constants $α$ and $β$.   Even with the hint I am lost. Of course $|h(z)|$ must attain its maximum when $|z|=1$, and $|h| \le |f|+|g|$, but that doesn't mean that $|f|+|g|$ couldn't attain its maximum inside the unit circle.  ny help?
93873
First definition of Riemann integrable function. Let $f:[a,b] \to \mathbb{R}$ be a bounded function and $P=\{x_0,x_1,\dots, x_n\}$ partition of $[a,b]$. Define $U(P,f):=\sum \limits_{i=1}^{n}M_i\Delta x_i$ and $L(P,f):=\sum \limits_{i=1}^{n}m_i\Delta x_i$ where $M_i=\sup\limits_{[x_{i-1},x_i]} f(x),\quad m_i=\inf\limits_{[x_{i-1},x_i]} f(x), \Delta x_i=x_i-x_{i-1}.$ Let $\inf \limits_{P}U(P,f)=I^*$ and $\sup \limits_{P}L(P,f)=I_*$. If $I^*=I_*=I$ then we called $f(x)$ is Riemann integrable function on $[a,b]$ with integral $I$.  Second definition of Riemann integrable function. Let $P=\{x_0,x_1,\dots, x_n\}$ is a partition of $[a,b]$ with $\xi_i\in [x_{i-1},x_i]$ and we define Riemann-integral sum $\sigma(P):=\sum \limits_{i=1}^{n}f(\xi_i)\Delta x_i$ and $\lVert P\rVert=\max\limits_{i}\Delta x_i$. If the following limit $\lim \limits_{\lVert P\rVert\to 0}\sigma(P)$ exists and has value $L$ we say that $f(x)$ is Riemann integrable function on $[a,b]$ with integral $L$.  The first definition is from Rudin's PMA book but in other books I met the second definition. But I can't prove the equivalence of these definitions for a couple days. Can anyone show to me a strict and rigorous proof? I would be very grateful for your help!  P.S. Happy New Year! :)
79998
I need to prove that     $f_1 \in \Theta(g_1) \land f_2 \in \Theta(g_2) \implies f_1 + f_2 \in \Theta(\max\{g_1, g_2\})$    is relevant, but I have a slightly different case, so I don't know how to translate it into this one, because I need to turn the $f$'s into $g$'s.  Am I allowed to say $\max\{f_1, f_2\} \in \Theta(\max\{g_1, g_2\})$?  Or even $\Theta(\max\{f_1, f_2\}) = \Theta(\max\{g_1, g_2\})$?
6281
Let $f,g,g´$ be continous on $[a,b]$ and $g$ monotone on $[a,b]$; then there exist $c\in (a,b)$ so that $$\int_{a}^{b}f(x)g(x)dx=g(a)\int_{a}^{c}f(x)dx+g(b)\int_{c}^{b}f(x)dx$$  Ineed to apply the first generalized mean value theorem for integrals i.e. $$\int_{a}^{b}f(x)g(x)dx=f(c)\int_{a}^{b}g(x)dx$$ to $g´$and $\int_{a}^{x}f(x)dx$ but I don´t know how please I would appreciate your help
32074
In an exercise session of an analysis course (which covers Riemann integration and differentiation in one dimension rigorously) the following question came up:  Suppose $f$ is strictly positive and integrable (on some compact interval $[a,b]$ on the real line). Can we show that $\int_a^b f &gt; 0$?  The proof by using measure theory and Lebesgue integration is easy, but also beyond the students at this point. So can one do without machinery such as sets of zero measure?    Tools in use: Mean value theorems, fundamental theorem of calculus, Riemann's condition for integrability (upper and lower integral within every epsilon of each other implies integrability), Riemann-Darboux integral, usual integration and differentiation techniques, as well as elementary real analysis in the epsilon-delta style.
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.
121099
I was reading the theory of Riemann integration when I cam across the following ,   If $f$ is bounded on $[a,b]$, and $P = \{x_0,x_1,x_2.......x_n\}$ is a partition of $[a,b]$, let $$M_j = \sup_{x_{j-1}\leq x\leq x_j}f(x)$$ The upper sum of f over P is $$S(P) = \sum_{j=1}^{n} M_j(x_j-x_{j-1})$$ and the upper integral of $f$ over $[a,b]$, denoted by $$\int_{a}^{b^-} f(x)dx$$ is the infimum of all upper sums.  The theorem similarly goes on to state the result for lower sums.  My doubt is : I do not understand how is $$\int_{a}^{b^-} f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partition P, then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P (but still being the lower limit does not prove that it is the infimum) and what about those partitions for which P itself is the refinement of?  How do I know that it will be a lower limit for those, let alone a infimum?
43973
I am working on a few examples where my author said this is true if $h$ is continuous. Can someone explain why? Just because the integral is $0$ doesn't mean the integrand is zero as well.   $\int_{a}^b h^2=0\implies h=0$?
74641
This is a step in a lecture note I'm reading. It should be simple because the author considers it obvious but I can't see it. What am I missing?  Suppose $U$ and $V$ are integrable over measure space $(\Omega,\mathcal{F},\mu)$. Claim: $$ \int_\Omega\sqrt{U^2+V^2}d\mu\geq\sqrt{\left(\int_\Omega Ud\mu\right)^2+\left(\int_\Omega Vd\mu\right)^2}. $$ Alternative phrasing: with $Z=U+iV$, claim that  $$ \int_\Omega|Z|d\mu\geq\left|\int_\Omega Zd\mu\right|. $$ Thank you for your time!
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.
122647
The problem is:  Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.  I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic.  Any help would be greatly appreciated, thanks!
19802
If $f(x)$ and $g(x)$ are two periodic functions, that is, $f(x+T_1)=f(x)$ and $g(x+T_2)=g(x)$ for every $x \in \Bbb R$. Now that $\displaystyle\lim_{x\to\infty}(f(x)-g(x))=0$.  Conjecture: $f(x) \equiv g(x)$.