An exercise in Real and Complex analysis (exercise 18, page 195,chapter 9)     Show that if a function $f$ on the real line $\mathbb{R}$ satisfies       \begin{equation*} f(x+y)=f(x)+f(y) \end{equation*}      and if $f$ possesses Lebesgue measure, then $f$ is continuous.   I can't do this question. Any advice or hints would be appreciated. Thank you in advance.
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A function $f:\Bbb R \to \Bbb R$ is additive and Lebesgue measurable. Prove that $f$ is continuous.  I know that on $\Bbb Q$, $f$ comes out to be linear. So, if $f$ is to be continuous then $f$ must be linear in $\Bbb R$. But, I'm stuck here.  If anyone can please help. Thanks in advance.
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Any help with this problem is appreciated. It comes up in the context of Lusin's theorem, where it was assumed to be true.     Suppose $E$ is a closed subset of $\mathbb{R}$ and $f:E\rightarrow \mathbb{R}$ is continuous. There exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x)=g(x)$ for $x \in E$ and $g$ satisfies $$\sup_{x \in \mathbb{R}} |g(x)| \leq \sup_{x \in E} |f(x)|\;.$$
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Some fellow classmates are preparing for a qualifying exam on real analysis, and asked me for help on the following question:  Let $ \ f:[0,1]^2\longrightarrow\mathbb{R}$ be such that: (i) $\ f(x,\cdot)$ is measurable for each fixed $x\in[0,1]$; (ii) $\ f(\cdot,y)$ is continuous for each fixed $y\in[0,1]$. Show that $\ f$ is measurable.  If we only assume that $\ f(\cdot,y)$ is measurable for each $y\in[0,1]$, rather than continuous, can we still conclude that $f$ is measurable?  I was able to come up with the following simple proof (at least I hope it is a proof) of the first statement using some standard arguments:  Proof. $ \ $ Define a sequence of functions $\{f_n:[0,1]^2\longrightarrow\mathbb{R}\}_{n\geq 1}$ by: $$ f_n(x,y)=f\bigg(\frac{i}{n},y\bigg), \qquad \text{for } x\in\bigg[\frac{i}{n},\frac{i+1}{n}\bigg], \ i=1,\cdots,n $$ Moreover, since $ \ f(\cdot,y)$ is continuous at $x=\tfrac{i}{n}$ for each $y$, for each $\epsilon&gt;0$ there is a $\delta_y&gt;0$ so that $$ \bigg| \ x-\frac{i}{n}\bigg|&lt;\delta_y \qquad\Rightarrow\qquad \bigg| \ f(x,y)-f\bigg(\frac{i}{n},y\bigg)\bigg|&lt;\epsilon; $$ therefore, for any $n&gt;\tfrac{1}{\delta_y}$ and any fixed $(x,y)\in[0,1]^2$ with $x\in\big[\tfrac{i}{n},\tfrac{i+1}{n}\big]$, we have $$ \bigg| \ x-\frac{i}{n}\bigg|\leq \frac{1}{n}&lt;\delta_y $$ and so $$ | \ f_n(x,y)-f(x,y)|=\bigg| \ f\bigg(\frac{i}{n},y\bigg)-f(x,y)\bigg|&lt;\epsilon. \tag{$\ast$} $$ In other words, $ \ f_n\longrightarrow f$ pointwise.  Furthermore, we clearly have that $$ A_n:=\{(x,y)\in[0,1]^2 \ | \ f_n(x,y)&gt;\alpha\} \qquad\qquad\qquad $$ $$ \qquad\qquad \ =\bigcup_{i=1}^{n-1}\bigg(\bigg[\frac{i}{n},\frac{i+1}{n}\bigg)\times \bigg\{y\in[0,1] \ \bigg| \ \ f\bigg(\frac{i}{n},y\bigg)&gt;\alpha \bigg\}\bigg). $$ Since $ \ f\big(\tfrac{i}{n},\cdot\big)$ is measurable for each $i=1,\cdots,n$, the latter expression is a union of measurable sets.  This means that $ \ A_n$ is measurable, and therefore $ \ f_n$ is measurable for each $n\geq 1$.  Finally, as the limit of measurable functions is measurable, we conclude that $ \ f$ is measurable. $\hspace{6.25in}\square$  I'm fairly certain that the above proof is correct (anyone care to confirm?); however, that still leaves the following:  Question:  If $ \ f$ is merely measurable in each variable separately, is it measurable?  The proof I gave above uses the continuity assumption in $(\ast)$, so obviously the same proof will not work in this case, but I cannot come up with any counterexamples.  Any suggestions?
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The question is the following: Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0$, $f'(x)$ exists. If $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists?  My intuition is that this does not have to hold in general. However, I keep finding counter examples such that $f'(0)$ exists where $\lim_{x \rightarrow 0} f'(x) = L$ does not exist, which is obviously not the original question.
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We have the function $f: \mathbb{R} \to \mathbb{R}$.     Question: Show that if the sets $\{x:f(x) &gt; a\}$ and $\{x:f(x) &lt; a\}$ are open for every $a\in\mathbb{R}$, then $f$ is continuous.   My attempt:$f$ is continuous if $A \subset \mathbb{R}$ is open, then $f^{-1}(A) \subset \mathbb{R}$ is open. I can show that the image of $\{x:f(x) &gt;a\}$ is open. Namely $\{f(x) &gt; a\} = (a,\infty)$ is open in $\mathbb{R}$. But this doesn't prove that for any open set $A \subset \mathbb{R}$, we have that $f^{-1}(A)\subset\mathbb{R}$ open as well.    Thanks in advance!
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I have read that if we let $(X, \tau_x)$ and $(Y, \tau_y)$ be topological spaces, and  $f: X \rightarrow Y$ is constant, then f is continuous. Can anyone please explain why? I assume we must look at the preimage of the open sets in $(Y, \tau_y)$..?  Thanks in advance
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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).
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Claim: If $f \in {\mathscr R[a,b]}$, then $f$ has infinitely many points of continuity.   1.) I read that it is a corollary of the Lebesgue integrability criterion.  Is it possible to prove the claim without invoking the concept of measure(or using less abstraction) ?    2.) Here is attempt: Given $\epsilon &gt; 0, \exists$ Partition, $P =  \left\{ x_0 =a,...,x_n =b \right\} $ of $[a,b]$ such that $\sum^n_{i=1} (M_i -m_i)\Delta x_i &lt; (b-a)\epsilon$, where  $M_i= \sup \left\{ f(x) : x \in \Delta x_i \right\}$ and $ m_i= \inf \left\{ f(x) : x \in \Delta x_i \right\} $  Let $(M_j -m_j)=\min \left\{ M_i -m_i : i=0,...,n \right\} $.  $ \implies (M_j-m_j)(b-a) \leq \sum^n_{i=1} (M_i -m_i)\Delta x_i &lt;  (b-a)\epsilon$ $ \implies (M_j-m_j)&lt;\epsilon. $  Let $ c\in (x_{j-1},x_j)$ and $\delta$ be any positive number such that $(c-\delta, c+\delta) \subseteq (x_{j-1}, x_j)$.   It follows that $ \left| f(x) - f(c) \right| &lt; \epsilon, $ whenever $ \left| x - c \right| &lt; \delta $.   Since $c$ is arbitrary and $[x_{j-1},x_j]$ is an interval,  there are infinitely many points of continuity.  There is something wrong with my proof since Thomae's function is a counterexample. Could anyone point out the mistakes in my proof? Thank you.
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Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y) = f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Show that if $f$ is continuous, then $f(x) = cx$ for all $x \in \mathbb{R}.$   I find it hard to use the definition of continuity to derive anything from the equation. I think we could write the definition of the limit in terms to get $$\forall \epsilon,\exists \delta \quad 0&lt;|x-a|&lt;\delta \quad \implies \quad |f(x)-L| &lt; \epsilon.$$ I am not sure where to go from here.
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Please help me to prove the following;  Let $ \ f:[0,1]^2\longrightarrow\mathbb{R}$ be such that: (i) $\ f(x,\cdot)$ is measurable for each fixed $x\in[0,1]$; (ii) $\ f(\cdot,y)$ is continuous for each fixed $y\in[0,1]$. Show that $\ f$ is measurable.
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I am struggling with an exercise in Topology. The exercise goes:  Let $Y$ be Hausdorff, and $X\subseteq Y$ an open subspace which is locally compact. Let $X^{+}$ denote the one-point compactification of $X$, and let $\infty$ denote the unique point in $X^{+} \backslash X .$ Define $f: Y \longrightarrow X^{+}$ by $$ f(y)=\left\{\begin{array}{ll} y &amp; \text { if } y \in X \\ \infty &amp; \text { else } \end{array}\right. $$ Show that $f$ is continuous.  I have a lot of definitions and lemmas about continuity, but none of them seems to fit with this problem. I am looking for a hint on how to get started with the proof. Thanks in advance.
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Suppose that $\|f\|_p &lt \infty$ for all $1\leq p &lt p'$, I want to know if the the following is true and in that case how to show it     $p \mapsto \|f\|_p$ is continuous on $[1,p')$   Or maybe we need to impose some more constraints such as finite measure space. In case of finite measure space, I tried to use .
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1) Construct a continuous function $f$ on $\mathbb{R}$ that is integrable on $\mathbb{R}$ but $\displaystyle\limsup_{x \to \infty} f(x)  = \infty$.  I took the function that is equal to $n$ on $[n, n + 1/n^{3})$ and made it continuous by saying that $f$ is the line segment joining $n$ and $n+1$ on $[n + 1/n^{3}, n+1)$. But I am failing to prove this integrable. For this, $\lim f(x) = \infty$ but how do you prove in general, if limit does not exist that $\limsup$ is infinity?  2) Prove that if $f$ is uniformly continuous and integrable on $\mathbb{R}$ we have $\displaystyle\lim_{|x| \to \infty} f(x) = 0$.  Any help is appreciated.  Thanks
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Given a continuous function $f:\mathbb{Q}\to\mathbb{Q}$ ,does there exist a continuous function  $g:\mathbb{R}\to\mathbb{R}$ such that   $g|_{\Bbb Q} = f$?  What I have no Idea about how to attempt this Question! Any suggestion will be very helpful.
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I'm annoyed by quite a simple problem in calculus (I apologize in advance if I'm not using adequate terms in English, I don't take the course in English nor am I a native speaker):  Let $f:\mathbb R\longrightarrow \mathbb R$ be a function which is continuous at $0$ and satisfies $f(x)=f(2x)\,\forall x\in\mathbb R$.  My question is how do I formally prove that for every $x&lt;0$ $f(x)=c_1$, and for every $x&gt;0$ $f(x)=c_2$, where $c_1,c_2\in \mathbb R$? After proving those two statements (or at least one of them WLOG) it would be quite easy to prove that $f(0)=c_1=c_2$ by the definition of continuity of $f(0)$ and thus $f(x)$ is constant. Those two statements are just so obvious to see that I can't think of any formal way of proving so without "cheating". Thanks in advance!
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This is part of a homework assignment for a real analysis course taught out of "Baby Rudin."  Just looking for a push in the right direction, not a full-blown solution.  We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero.  The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant.  I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers.  Is this an approach worth taking?
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My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:  $$|f(x)-f(y)|\geq k|x-y|$$  Then $f$ is bijective and  its inverse is continuous.  Well, there's a Theorem , Invariance of domain, that says   "If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".  But I'm not knowing how to proceed...need a clue...Thanks for attention!!!