Let $f$ be a real-valued function on $A$ in $E^n$. Show that is $∫_Af$ exists, then so does $∫_A|f|$.   I have been given an outline of what I need to do, however, I have been unable to actually formulate it into a rigorous proof. Here is the outline:   By definition, Riemann integrable functions are bounded. Note that for every partition $P$ and every cell $c$ in $P$, $\sup_c|f|−\inf_c|f|⩽\sup_c f−\inf_c f$ hence $U(|f|,P)−L(|f|,P)⩽U(f,P)−L(f,P)$, this should show that $|f|$ is Riemann integrable.   EDIT: The hint above was given .
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Let $f$ be a real-valued function on $A$ in $E^n$. Show that is $\int_A f$ exists, then so does $\int_A |f|$ and $\left | \int_A f\right | \le \int_A |f|$.   I am trying to finish Rosenlicht's Introduction to Analysis and the last chapter is awful. But I do want to finish it because I have come this far. Could someone show me how to do this, so I can at least have some example problems to understand the material. This chapter (Multiple Integrals) really just left more confused than before I started reading the chapter.  Thank you in advance.
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Assume $f$ is Lebesgue measurable and bounded on $[0,1]$ and that $$\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx\to0\;\text{as}\;h\to0.$$ Show that $f$ is constant for almost every $x$ in $[0,1]$.  My question relates to my proposed proof below.  At no point did I rely on the boundedness of $f$ on $[0,1]$, only its measurability and so I suspect my reasoning is faulty and would appreciate somebody that could (in their answer) scrutinize and correct my proof.  (Proposed) Proof. Assume first that $h\to0^{+}$ and suppose that we may interchange the two limits.  Then since the difference quotients are non-negative, we conclude $f'(x)$ exists and is equal to $0$ for almost every $x$ in $[0,1]$, and the claim follows. (As pointed out, we don't necessarily know the limit of the integrands exists even when we assume the interchange is valid, although this is probably not difficult to reason one's way out of -- more damning is below.)  The goal quickly turns to justifying this interchange.  To that end, the presumed existence of the above limit implies that for any $0&lt;M&lt;\infty$ there is some positive real number $h(M)$ such that for any $h&lt;h(M)$ we have $$\left|\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx\right|&lt;M.$$ Since $m([0,1])=1$ (in particular, we are integrating over a set of finite measure) and the difference quotients are non-negative, we conclude that $$\frac{|f(x+h)-f(x)|}{h}&lt;M\;\text{a.e.}\;x.$$ Since this bound is independent of $h$ once $h&lt;h(M)$, it follows that at some point along the sequence each of the difference quotients is pointwise a.e. dominated by $M$, which is integrable on any set of finite measure.  Hence the interchange of limits discussed above is valid, and with the case of $h\to0^{-}$ being handled similarly, the claim is proved.   (As the comment points out, we cannot conclude anything about the upper (or lower) bound of a function based off the value of its integral, even if the integration is taken over a set of finite measure and the integrand in question is nonnegative; approximations to the identity serve or otherwise spikey functions on sets of small measure as counter-examples.)    Edit. Fatou's lemma implies $$\int_{0}^{1}\liminf_{h\to0}\frac{|f(x+h)-f(x)|}{h}\;dx\leq\lim_{h\to0}\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx=0.$$  Hence $$\liminf_{h\to0}\frac{|f(x+h)-f(x)|}{h}=0\;\text{a.e.}\;x.$$ (The $\liminf$ of course exists without further justification.)  We now need a way to obtain the same for $\limsup_{h\to0}\frac{|f(x+h)-f(x)|}{h}.$  This could be accomplished by showing the limit exists under the assumptions given, in which case the $\liminf$ is equal to the $\limsup$.  Any ideas how to execute this, or otherwise approach the problem?  I'm wondering too if my original argument above could be modified by taking into account the assumed boundedness of $f$ in order to exclude the possibility that its difference quotients form a "spikey" sequence of integrands.    Edit. Here is another approach I am considering by posting this question   Peter Tamaroff mentioned the type-writer sequence being a counter-example to the post linked.
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Let $f \in \mathbf R[x]$  and for every $k \in \mathbf Z$, $f(k)\in \mathbf Z$. Does this imply that $f \in \mathbf Z[x]$?   I tried to find a counter example, but my intuition tells me it's true and I don't know how to prove the statement either.
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Let $\mathcal{F}$ be a family of analytic functions in a domain $G$. Denote by $\mathcal{F}':=\{ f' \mid f \in \mathcal{F}\}$. It does not suffice to know that $\mathcal{F}'$ is normal to ensure that $\mathcal{F}$ is normal (namely take $\mathcal{F} = \{ n \mid n \in \mathbb{N}\}$ as a counterexample) but someone assured me that if you assume at some fixed point $z_0 \in G$, if you know that $f(z_0) = 0$ for all $f \in \mathcal{F}$ and that $\mathcal{F}'$ is normal, then $\mathcal{F}$ is normal. But I can't see how to do it.   A friend tried something with line integrals to show that the $f$ are locally bounded on compact subsets, but he only considered compact sets that were the image of a path, which I don't think is sufficient. Which characterization of normality should I try (local boundedness? convergence in the space of functions?)? Does anyone have an idea?
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Question:  a. Find an example for a non-negative and continuous function s.t. $\int _0^\infty f(x)dx$ is finite but the following limit doesn't exist: $\lim_{x\to \infty} f(x)$.  b. Is it possible that $\int _0^\infty f(x)dx$ is bounded but $f(x)$ is not bounded?  What we did with A: We suggested the function $|sin(x^2)|$ which has periods that get smaller and smaller until they no longer imply on the sum. But since it's a trig function it doesn't have a determinate limit. Wolfram didn't have the integral value for that func, and we were wondering if it really converges and if it is really a good example.  What we did with B: We thought about the function: $f(x)=  { x\in \Bbb N },{e^{-x} \notin \Bbb N }$ and we had a disagreement whether f(x) is integrable. I said no because similarly to Dirichlet function, one sum's limit will be 0 while the other one's will be infinite. My partner said that if I find a $\delta&lt;1$ then the definition of Riemann's integral does hold and so this integral is equal to that of $f(x)=e^{-x}$ She was trying to say this functions integral will be 0, but still it won't be bounded. I disagreed. Who's right?
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Suppose that $f$ is holomorphic on $\mathbb C$ and that $\Re(f(z))\ge 0$ for all $z$. Show that $f$ is constant. [Hint: consider $e^{−f(z)}$.]   My thoughts: If $\Re(f(z))\ge 0 $ holds, then $e^{−f(z)}$ is a bounded holomorphic function (do I need to prove this or is it obvious?) So then by Liouville's Theorem $e^{−f(z)}$ is constant.   But then I'm not sure how to rigorously go from this $\exp(−f(z))$ back to $f(z)$.   Could anyone help me piece this together please?  Thanks
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I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that  $$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$  I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!
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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.
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I'm having a little trouble with this homework problem:     Suppose $\mu(X)&lt;\infty$, $f_n\in L^1$, $f_n\to f$ a.e., and there exists $p&gt;1$ and a constant $C&gt;0$ such that $$\|f_n\|_p\leq C$$ for all $n.$ Prove that $f_n\to f$ in $L^p$.   Here's what I've done so far.  Without loss of generality, suppose that $\mu(X)=1$. Since $\|f_n\|_p\leq C$, we have that $f_n\in L^p$ for all $n$. Furthermore, since $|f_n|^p\to |f|^p$, we have that $$\int |f|^p\leq \liminf \int|f_n|^p\leq C^p$$ by Fatou's lemma, so $f\in L^p$ as well.    Since $|f_n- f|\to 0$, we have that $|f_n-f|^p\to 0$.   I am thinking of applying the Dominated convergence theorem to $|f_n-f|^p$, but I cannot think of a bound.
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I am thinking about this since couple hour. Is a continuous and bounded function $f:\mathbb R\to\mathbb R$ uniform continuous too? I didn't found a counter example and thus I tried to prove this like this: let $\epsilon&gt;0$ and $a\in\mathbb R$. By continuity, there is a $\delta_a&gt;0$ such that $$d(f(x),f(y))&lt;d(f(x),f(a))+d(f(y),f(a))&lt;\epsilon$$ if $d(x,y)\leq d(x,a)+d(y,a)&lt;\delta_a.$ Then, if $\inf_a\delta_a&gt;0$ I set $\delta=\inf_a\delta_a$ and we are done since $d(f(x),f(y))&lt;\epsilon$ if $d(x,y)&lt;\delta,$ but I didn't success to prove that $\delta\neq 0$.
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Let $f \in C[0,1]$ and $f(0)=f(1)$.   How do we prove $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$?  In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$.  For any other non-zero real $r$ (i.e not of the form $\frac{1}{n}$), there is a continuous function $f \in C[0,1]$, such that $f(0) = f(1)$ and $f(a) \neq f(a+r)$ for any $a$.  This is called the Universal Chord Theorem and is due to Paul Levy.  Note: the accepted answer answers only the first question, so please read the other answers too, and also this answer by Arturo to a different question:     This is being repurposed in an effort to cut down on duplicates, see here: .  and here: .
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The problem is in Marsden - Elementary Classical Analysis 2nd ed. Ch.5.  My approach was   Condition : $f_k, f(\in\mathcal C) : A\to N$, $f_k \to f \ \text{(pointwise)}$ and $A$ is compact. Since $A$ is compact, the range of $f_k, f$ are all compact, so bounded. Since $f_k, f\in\mathcal C_b$, if $\|f_k-f\|\to 0$ in $\mathcal C_b$ then $f_k\to f \ \text {(uniformly)}$.   when the norm of $\mathcal C_b$ is defined by $\|f\|=\sup\{\|f(x)\||x\in A\}$.  I got stucked in 3, showing $\|f_k-f\|\to 0$. I don't even know the proposition is true. Can anyone give me some hint to solve this or a counterexample for this?
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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.
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Let $f:[0,1]\to \mathbb{R}$ be differentiable on [0,1]. We know that the derivative of $f$ denoted by $f'$ may not be Riemann integrable on [0,1].  We know that a necessary condition for $f'$ to be Riemann integrable is boundedness of $f'$ on $[0,1]$.  My question:   Is boundedness of $f'$ on [0,1] also a sufficient condition? What are other sufficient conditions of $f'$ to be Riemann integratble on [0,1]?
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I am trying to prove the following:  If $A\subset\mathbb R$ is bounded and $\,f:A\to \mathbb R\,$ is uniformly continuous, then $f[A]$ is bounded.  Could you check my proof?  Let $A \subseteq [-K,K]\subseteq \mathbb R$. Let $\varepsilon = 1$. If $f$ is uniformly continuous there is $\delta$ wuth $|x-y| &lt;\delta$ imply that $|f(x)-f(y)|&lt;1$ for all $x,y\in A$. Let $a \in A$. Then because $A$ is bounded there is a finite number of balls $B(a_n,\delta)$ that cover $A$. Let the number be $N$. Then $f(a)-N \le f(x) \le f(a) + N$ for all $x\in A$.
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Can you give me an example of a bounded f:[0,1]-> R , which has a primitive but it isn't Riemann-integrable? It must be on a closed interval from zero to one.
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Prove that if $f$ is Riemann integrable on $[a,b]$ and $0&lt;\rho \le f(x)$ for all $x \in [a,b]$ then $1/f$ is Riemann integrable on $[a,b]$.