Suppose we have a set $A\subset\mathbb {R}$ and let $f\in\mathcal{B}(A)$ and $g\in\mathcal{B}_b(A)$ (Borel function on $A$ and bounded Borel function on $A$, resp.) Is it possible to approximate $f$ and/or $g$ by continuous functions on $A$ in a certain sense (pointwise, uniform, etc.)? So yes, how to prove that, or what are the references? So no, what is a counterexample?  Thanks a lot.
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This is somewhat linked to a  which was looking to see if a proof of mine regarding the Dirichlet function was correct (it wasn't). I now have an answer to the question which can be answered without using directly using the Baire category theorem or the likes; as it is sufficiently different to my original approach, I feel that a new question would be the best way of going about this.  The question is to     Construct a function $f:[0,1] \rightarrow \mathbb{R}$ which is not the pointwise limit of any sequence $(f_n)$ of continuous functions   to which I will have an answer below (while I didn't come up with much of this myself, I feel it's an interesting result to discuss).  Finally, as a warning to those who are seeing this as a result of searching for answers to their example sheet/homework questions, please have a think about the question before reading the answer below.
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Find all the continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that if $x-y\in \mathbb{Q}$ then $f(x)-f(y)\in \mathbb{Q}$.   Can somebody give me some hint to solve this problem. I think that the functions are $f(x)=mx+b$ where $m\in\mathbb{Q}$.  Thanks!
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Suppose that $S$ is a closed subset of $\mathbb R^d$ and that $f$ is a continuous real-valued function with domain $S$. Show that, for every real number $a$, the set {$x ∈ S | f(x) ≤ a$} is a closed subset of $\mathbb R^d$    I don't know how to start this. Do we use the fact that If $S$ is closed and bounded every sequence in $S$ has a subsequence that converges to a point in $S$?
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Let $f:\mathbb{R^n}\to\mathbb{R^m}$ be a function such that the image of any closed bounded set is closed and bounded. Must $f$ be continuous?
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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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I was trying to prove an important theorem concerning continuous functions, namely that the product of two continuous functions is continuous. I am nearly at the end of the proof but I do not understand the last step my teacher made... I'm sure some of you is able to help me out :) Thanks in advance!  Theorem  Let $f, g\colon \Bbb R \to\Bbb R $ be continuous. Then $F\colon \Bbb R\to\Bbb R$ defined by $F(x) = f(x)g(x)$ is continuous.  Proof  Let $f, g\colon \Bbb R \to\Bbb R $ be given such that $f$ and $g$ are continuous.  Let $F\colon \Bbb R\to\Bbb R$ be defined by $F(x) = f(x)g(x)$. We want to show: $F$ is continuous, that is, for all $a \in\Bbb R$, for every $\epsilon &gt; 0$, there exists some $\delta &gt; 0$ such that for all $x \in \Bbb R$ with $|x - a| &lt; \delta$, $|F(x)-F(a)| &lt; \epsilon$.  Let $a \in\Bbb R$ be given. Let $\epsilon &gt; 0$ be given.   Let $\delta_f &gt; 0$ be such that for all $x \in\Bbb R$ with $|x-a| &lt; \delta_f$, $|f(x)-f(a)| &lt; \frac\epsilon{2(|g(a)| + \epsilon)}$. Such a $\delta_f$ exists since $f$ is continuous.  Let $\delta_g &gt; 0$ be such that for all $x \in\Bbb R$ with $|x-a| &lt; \delta_g$, $|g(x)-g(a)| &lt; \frac\epsilon{ 2|f(a)| + 1}$. Such a $\delta_g$ exists since $g$ is continuous.  Let $\delta_3 &gt; 0$ be such that for all $x\in\Bbb R$ with $|x-a| &lt; \delta_3$, $|g(x)-g(a)| &lt; \epsilon$, which implies $|g(x)| &lt; |g(a)| + \epsilon$. Such a $\delta_3$ exists since $g$ is continuous.  Choose $\delta = \min\{\delta_f, \delta_g, \delta_3\}$.  Then for all $x \in\Bbb R$ with $|x-a| &lt; \delta$:  $$\begin{align}|F(x) - F(a)| &amp;= |f(x)g(x) - f(a)g(a)| \\ &amp;= |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)| \\ &amp;\le |f(x)g(x) - f(a)g(x)| + |f(a)g(x) - f(a)g(a)|\\ &amp;= |(g(x)[f(x) - f(a)]| + |f(a)[g(x) - g(a)]|\\&amp;= |g(x)| * |f(x)-f(a)| + |f(a)| * |g(x) - g(a)| \\ &amp;&lt; [|g(x)| + \epsilon] \cdot \frac \epsilon {2(|g(a)| + \epsilon)} + |f(a)| \cdot |g(x) - g(a)| \\ &amp;= \frac\epsilon{2 + |f(a)| \cdot |g(x) - g(a)|}\end{align}$$  Problem  Now we need $|f(a)| \cdot |g(x) - g(a)|$ to be equal to $\frac \epsilon 2$ so that we have $\frac\epsilon2+\frac\epsilon 2= \epsilon$ (which completes the proof). The problem is I do not see why/how $|f(a)| \cdot |g(x) - g(a)|$ is equal to $\frac\epsilon2$. Or is there maybe an error somewhere?
16861
Let $m$ be Lebesgue measure on $\mathbb R$ and $\mathcal{F}$ a countable set of Lebesgue measurable functions defined on Lebesgue measurable set $E\subset\mathbb R$ with $m(E)&lt;\infty$. Assume that for each $x\in E$, the set $\{|f(x)|:f\in\mathcal F\}$ is a bounded subset of $\mathbb R$. Show that for all $\epsilon&gt;0$, there is a closed set $F\subset E$ and a number $M&lt;\infty$ such that $m(E\setminus F)&lt;\epsilon$ and $|f(x)|\le M$ for all $x\in F$ and all $f\in\mathcal F$.     My attempt:  By , for each $f_n\in\mathcal F$, there exists a closed subset of $F_n$ such that $f_n$ is continuous on $F_n$ with $m(E\setminus {F_n})&lt;\epsilon/2^n$. Then for every $f\in\mathcal F$, $f$ is continuous on the closed set $F=\bigcap_{n=1}^\infty F_n$ with $m(E\setminus F )&lt;\epsilon$.   Now how to prove that it is uniformly bounded on $F$?
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Let $f$ be a real-valued Lebesgue measurable function on $\mathbb{R}$. Prove that there exist Borel measurable functions $g$ and $h$ such that $g(x)=h(x)$ almost everywhere and $g(x)\le f(x) \le h(x)$ for every $x \in \mathbb R$. I know that $f$ is measurable since there exists a sequence of simple function that converges to $f$. I have no further idea how to tackle this problem.
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Let $f$ be a real valued and continuous function over $[a,b]$ with $a &lt;b$ such that $f(a)=f(b)$ , My question here is :     Question (claim): How to show that : there  exists  $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.   Note: I have tried intermediate values but I didn't succeed.
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Let $D = \{(x,y)\in \mathbb{R}^2: x^2 + y^2 \leq 1 \}$. Let $A \subset \mathrm{int}D$. Let $A$ be connected and compact and let $D \setminus A$ be connected. Let $f:A \longrightarrow A$ be a continuous function and let $g:D \longrightarrow D$ be also continuous function such that $g_{|A} = f$.$\\$ Does it imply that there exists $x \in \mathrm{int}D$ such that $g(x)=x$?
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I'd like to prove the following:  If $f_j$ are continuous functions on a compact set $K$, and $f_{1}(x) \leq f_{2}(x) \leq \dots$ for all $x \in K$, and the $f_j$ converge pointwise to a continuous function $f$ on $K$ then in fact the $f_j$ converge uniformly to $f$ on $K$.  Attempt:  Let $g_{j}(x) = f(x) - f_{j}(x)$ for all $j$. Then, since $f_j \rightarrow f$ pointwise, we see $g_j \rightarrow 0$ pointwise.  Now, let $\varepsilon &gt; 0$ . And examine { $x \in K : g_{j}(x) &lt; \varepsilon$ }.  I've been told that the next step should be to  show that { $x \in K : g_{j}(x) &lt; \varepsilon$ } is equal to the intersection of $K$ with some open set $U_j$. But I'm not certain why this is true?   Advice? Insight?
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A problem that I have been thinking: For what kind of functions $f:[a,b]\to \mathbb R$ is  $x\mapsto\sup_{t\in [a,x]}f(t)$ continuous? This is not true for bounded functions, take for example $f(x)=0$ in $[0,1]$ and $f(x)=1$ in $(1,2]$. It is however trivially true for continuous monotone functions. Assuming mere continuity for $f$ I can show right continuity of $\max_{t\in[a,x]}f(t)$:   For $\epsilon$ and some $\delta&gt;0$,  $$x\in [x_0-\delta,x_0+\delta]\implies f(x_0)-\epsilon&lt;f(x)&lt;f(x_0)+\epsilon$$ If $x\in [x_0,x_0+\delta]$ then $$\max_{t\in [a,x]}f(t)=\max\left\{\max_{t\in [a,x_0]}f(t),\max_{t\in [x_0,x]}f(t)\right\}$$ If $max_{t\in [a,x_0]}f(t)\ge \max_{t\in [x_0,x]}f(t)$ then $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|=0&lt; \epsilon$$ Otherwise, $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|\le \max_{t\in [x_0,x]}f(t)-\max_{t\in [a,x_0]}f(t)&lt;f(x_0)+\epsilon-\max_{t\in [a,x_0]}f(t)\le \epsilon$$ which proves right continuity. When I tried to prove left continuity I was stuck on showing $$\max_{t\in [x,x_0]}f(t)-\max_{t\in [a,x]}f(t)&lt;\epsilon$$ Can this be circumvented?
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We know that if $ f: E \to \mathbb{R} $ is a Lebesgue-measurable function and $ g: \mathbb{R} \to \mathbb{R} $ is a continuous function, then $ g \circ f $ is Lebesgue-measurable. Can one replace the continuous function $ g $ by a Lebesgue-measurable function without affecting the validity of the previous result?
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I'm looking for subset $A$ of $\mathbb R$ such that $A$ is a Borel set but $A$ is neither $F_\sigma$ nor $G_\delta$.
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Suppose consider the space (R,d), where d is the absolute value metric, and f is a continuous function from R to R. Let A be a subset of R. Then is  Cl(f(A)) a subset of f(Cl(A))? Prove or provide counterexample. If yes, can you extend in for any arbitrary metric spaces I am very new to metric space and topology, so any help will be appreciated.
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I have to prove the following:  Let $f: \mathbb{R^2}\to \mathbb{R}$ such that $f_x:y\to f(x,y)$ is Borel measurable for all $x\in\mathbb{R}$ and that $f^y:x\to f(x,y)$ is continuous for all $y\in\mathbb{R}$. Prove that $f$ is Borel measurable.  What I have tried to do is to find a sequence of functions $f_n(x,y)$ s.t for a fixed $y$ $f_n(.,y)$ is a linear approximation of $f(.,y)$..
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Possible Duplicate:          Let $f, g:\Bbb{R}\to\Bbb{R}$ to be continuous functions such that $f(x)=g(x)\text{ for all rational numbers}\,x\in\Bbb{Q}$. Does it follow that $f(x)=g(x)$ for all real numbers $x$?  Here is what I think:  f continuous when $\lim\limits_{x\to x_0}f(x)=f(x_0)$ and $\lim\limits_{x\to x_0}g(x)=g(x_0)$  So it does not neccesarily mean that $f(x)=g(x)$ when x is irrational. So I can pick a function f so that   $f(x) = \begin{cases} g(x)  &amp; \text{if $x\in\Bbb{Q}$} \\ x &amp; \text{if $x\in\Bbb{R}\setminus \Bbb{Q}$}  \\ \end{cases}   $