Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a monotone increasing function. We know from a general fact that the set of discontinuities of $f$ is countable. Denote the set of discontinuities by $D$.  I want to construct a continuous function $g$ such that there exists a function $h$ defined on $D$ with $f(b)-f(a)=g(b)-g(a)+\sum_{x\in (a,b)\cap D}h(x)$.  The problem is that there can be a countable number of discontinuities, and it is not necessary that we can order them as $\cdots&lt;x_{-2}&lt;x_{-1}&lt;x_0&lt;x_1&lt;x_2&lt;\cdots$. How can we construct this function $g$?
122184
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a monotone increasing function. We know from a general fact that the set of discontinuities of $f$ is countable. Denote the set of discontinuities by $D$.  Define $g:\mathbb{R}\rightarrow\mathbb{R}$ as follows: $$g(x)=f(x^-)-\sum_{y\in D \text{ and }y&lt;x}(f(y^+)-f(y^-))$$ It should be relatively clear that $g$ is monotone increasing. But is $g$ necessarily continuous?
58828
Let $a,b\in \mathbb{R}: a&lt;b$. Let $f:[a,b]\to \mathbb{R}$, $f$ is monotone increasing. Let $F:[a,b]\to \mathbb{R}, x \mapsto \int_{[a,x]} f$.  Show that if $F$ is differentiable at $x_0 \in [a,b]$, then $f$ is continuous at $x_0$.  If f was not continuous, then the right and left limits would not match. Because the function jumps, its measure has to jump too, therefore integral would be discontinuous and therefore non differentiable. However I do not know how to formalize this intiuition.
89560
I am having difficulties showing that this is true. How would you prove this statement?  Given that the function $h: \mathbb{R}\rightarrow\mathbb{R}$ is increasing. Show that $h$ is continuous if and only if  $h(\mathbb{R})$ is connected.
55720
Do there exist functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y),$ but which aren't linear? I bet you they exist, but I can't think of any examples.  Furthermore, what hypotheses do we need to put on $f$ before no such functions exist? I feel continuity should be enough.
3731
Given that a function $g(x)$ is a monotone increasing function. Its domain is an interval [c,d]. There are two sets $a_j \in A, j= 1,...,J$ and $b_k \in B,k=1,...,K$  on this interval, which satisfy $c_z \in C = B$\A and $c_z \geq a_j$ for all $a_j \in A$. For example, the interval $[c,d]=[0,8]$,$A=\{0,1,2,3,4,5\}$ and $B=\{0,1,2,3,4,5,6,7\}$. In this case, does the following inequality hold? If yes, could you give me some hints to prove that?  $$\frac{1}{K} \sum_{k=1}^{K}g(b_k) \geq \frac{1}{J} \sum_{j=1}^{J}g(a_j)$$
28845
Show that there exists a not decreasing function $f:(a,b)\rightarrow\mathbb{R}$ continuous on $(a,b)\setminus D$ and discontinuous on $D$ where $D$ is a countably infinite subset of $(a,b)$.  This is an exercise that I tried very last semester but my weak point is to get examples of functions. Can anyone help me with this?  *Moreover, one could indicate something for me to improve my skill with examples?
53131
My teacher showed us this function and told us it was continuous at all non-$\mathbb{Q}$ points:  $$ f(x) = \begin{cases}   x &amp; \text{ if } x\in\mathbb{Q} \\\\   0 &amp; \text{ if }x\notin\mathbb{Q} \end{cases} $$  However, Wolfram MathWorld says , which is very similar, is discontinuous at all points. Why is one continuous and the other not?
46933
I tried to solve this problem but I could not continue  Let $f: [a, b​​]\to \mathbb{R}$ be a continuous function.  Prove that if $x_1, x_2,\ldots,x_n  \in (a,b​)$, then exists $x_0 \in (a, b​​)$ such that $$f (x_0) =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}$$  the function is continuous, so $\forall\ x_1, x_2,\ldots,x_n  \in(a, b​​)$ $$\begin{align*} \lim_{x \to \ x_1}f(x)&amp;=f (x_1)=L_1\\ \lim_{x \to \ x_2}f(x)&amp;=f (x_2)=L_2\\ \lim_{x \to \ x_3}f(x)&amp;=f (x_3)=L_3\\ &amp;\vdots\\ \lim_{x \to \ x_n}f(x)&amp;=f (x_n)=L_n\\ \end{align*}$$  We have that:  $$\lim_{x \to \ x_0}f(x)=f (x_0)=\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}= \frac{1}{n}\sum_{j=1}^n f(x_j)$$ Now $$\min\left \{ L_1,L_2,L_3,\cdots,L_n \right \}\le\frac{1}{n}\sum_{j=1}^n f(x_j)\le\max\left \{ L_1,L_2,L_3,\cdots,L_n \right \}$$  and at this point I'm lost ...
33321
The following is not a homework problem. I am doing it for self study.      Prove that any continuous function from $\mathbb{R}$ to $\mathbb{Q}$ is constant.    Here is my proof:     Let $f:\mathbb{R}\rightarrow \mathbb{Q}$ be such a function.    ? We first show that $\mathbb{Q}$ is disconnected. Let $p$ be an irrational number. Then we can write $\mathbb{Q}$ as  $$ \big( \mathbb{Q} \cap (-\infty,p) \big) \cup \big( \mathbb{Q} \cap (p,\infty) \big),  $$  which is a disconnection of $\mathbb{Q}$.      Moreover, $\mathbb{Q}$ is also totally disconnected, as any open subset $U$ of $\mathbb{Q}$ must contain two rational numbers and there is always an irrational number between the two, and we can use that rational number to create a disconnection. Therefore the only connected subsets of $\mathbb{Q}$ are the singleton sets $\{q\}$, where $q \in \mathbb{Q}$, which are closed.       We will show that $f(\mathbb{R})$ must be connected. Assume not, then    $$ f(\mathbb{R})= U\cup V, $$   where $U$ and $V$ are nonempty open subsets of $\mathbb{Q}$ such that    $$ U \cap V= \emptyset .$$    This implies   $$ \mathbb{R} = f^{-1} \big( U \cup V \big) = f^{-1}(U)\cup f^{-1}(V),  $$   where $f^{-1}(U),f^{-1}(V)$ are nonempty, nonintersecting subsets of $\mathbb{R}$. This, however, is a contradiction, as $\mathbb{R}$ is connected. Therefore no such set $U$ and $V$ can exist.       As we have already shown, the only connected subsets of $\mathbb{Q}$ are the singleton sets $\{q\}$, so $f(\mathbb{R})$ must be such a set.   Something about this does not seem quite satisfactory, as if I am missing something. Could anyone tell me a flaw in my logic? Also, is there a more satisfactory way to prove this theorem?
44696
Find functions $f,g : \mathbb R \longrightarrow \mathbb R $ and $a,b,c \in \mathbb R $ such that  $$ \lim_{x \rightarrow a} f(x) = b \quad \text{and} \quad \lim_{y \rightarrow b} g(y) = c \quad \text{but}\quad \lim_{x \rightarrow a} g(f(x)) \neq c$$  I tried to use non continous functions but i didnt manage to get a any useful results.
64888
If $f,g$ are continuous at $a$, show that $h(x)=\max\{f(x),g(x)\}$ and $k(x)=\min\{f(x),g(x)\}$ are also continuous at $a$. Here is my attempt at a proof. It feels very elaborate and I am not sure if it is correct. Can someone please point out any mistakes or places where I may improve. Thanks!  By the definition of continuity at $a$ of $f,g$ we have that $\lim\limits_{x\to a}f(x)=f(a)$ and $\lim\limits_{x\to a}g(x)=g(a)$. Suppose that $f(a)&gt;g(a)$. Then there exists a $\delta&gt;0$ such that $f&gt;g$ for all $x$ satisfying $|x-a|&lt;\delta$. Then for $x$ satisfying $|x-a|&lt;\delta$ we have $h(x)=f(x)$ and $k(x)=g(x)$ and thus $h$ and $k$ are continuous at $a$ because $f$ and $g$ are continuous at $a$. Now if $f(a)&lt;g(a)$ we simply relabel $f=\tilde{g}$ and $g=\tilde{f}$, so that $\tilde{f}(a)&gt;\tilde{g}(a)$ and we apply the previous result to show that again $h$ and $k$ are continuous at $a$.\  Now if $f(a)=g(a)$ then we can distinguish three cases. $f\geq g$ in a small neighborhood about $a$, $f\leq g$  in a small neighborhood about $a$ or $f\geq g$ on one side of $a$ and $f\leq g$ on the other side. In the first case we assume that $f\geq g$ in some small neighborhood about $a$. Then, since $f(a)=g(a)$ we have $h(x) = f(x)$ in this neighborhood and $k(x) = g(x)$ and again we see that $h$ and $k$ are continuous at $a$ due to the continuity of $f$ and $g$ at $a$. Similarly if $f\leq g$ in a neighborhood around $a$ then $h(x)=g(x)$ and $k(x)=f(x)$ in this neighborhood and $h$ and $k$ are continuous at $a$.   Suppose that just to the left of $a$ we have $f\geq g$ and just to the right of $a$ we have $f\leq g$. Then $\lim\limits_{x\to a^-}h(x)=\lim\limits_{x\to a^-}f(x)=f(a)=h(a)=g(a)=\lim\limits_{x\to a^+}g(x)=\lim\limits_{x\to a^+}h(x)$. Similarly $\lim\limits_{x\to a^-}k(x)=\lim\limits_{x\to a^-}g(x)=g(a)=k(a)=f(a)=\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^+}k(x)$. So $\lim\limits_{x\to a}h(x)=h(a)$ and $\lim\limits_{x\to a}k(x)=k(a)$ which means that $h$ and $k$ are continuous at $a$. Lastly if $f\leq g$ just to the left of $a$ and $f\geq g$ just to the right of $a$ we can relabel $f=\tilde{g}$ and $g=\tilde{f}$ and apply the last result to show $h$ and $k$ are continuous at $a$.   Again, please point out any mistakes I may have made. Thanks!!
85253
I am working through a bank of previous exams and couldn't figure a problem out to my satisfaction.     Let $f(x) : \mathbb{R} \to \mathbb{R}\,$ be a continuous function.         Show that $f$ can have at most countably many strict local maxima.   Assume that $f$ is not monotone on any interval. Then show that the local   maxima of $f$ are dense in   $\mathbb{R}$.
79979
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?
27267
Suppose that the function $f:[a,b]\rightarrow \mathbb{R}$ is continuous. For a natural number $k$, let $x_1,\cdots,x_k$ be points in $[a,b]$. Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$.    This can be proved by showing   $\hspace{130pt}f(a)&lt;\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}&lt;f(b)$  and using the intermediate value theorem to say there exists a point $z\in [a,b]$ such that the statement is true.  I'm not sure how to show this though. Any suggestions?
49576
I have really no idea about this:  Problem: Show that there exists a function $f:[0,1]\rightarrow\mathbb{R}$ such that:  $f$ is discontinuous in all $x\in \mathbb Q$. $f$ is increasing in $[0,1]$. $f$ is integrable.   EDIT: Sorry, it is not discontinuous in all $x\in \mathbb R \setminus \mathbb Q$, just in $\mathbb Q$.
84042
As a homework question, we got asked the following:     Construct a function $f:[0,1] \rightarrow \mathbb{R}$ which is not the pointwise limit of any sequence of continuous functions   Thinking about it, a function which is nowhere continuous should be sufficient; thus I decided to try to use the indicator function on the rationals, $1_\mathbb{Q}$. Basically, I was just wondering if my proof was alright or not:  Suppose $(f_n)$ is a sequence of continuous functions converging pointwise to $1_\mathbb{Q}$; thus for any $\epsilon &gt; 0$ we have that for all $x$ we can find a $N$ such that if $n \geq N$ then $| f_n(x) - 1_\mathbb{Q} (x) | &lt; \epsilon$. In particular, we have for some $x_1 \in \mathbb{Q}, x_2 \notin \mathbb{Q}$ a $N_1$ and $ N_2$ respectively such that   if $ n \geq N_1$ then $ | f_n(x_1) - 1 | &lt; \epsilon$ if $ n \geq N_2$ then $ | f_n(x_2) | &lt; \epsilon$   Now, as $f_k$ is continuous, for any $\epsilon &gt; 0$ we have that for all $x$ there exists a $\delta &gt; 0$ such that if $ | x- y | &lt; \delta$ then $|f_k(x) - f_k(y) | &lt; \epsilon $. However, if we pick an arbitrary $x \in \mathbb{Q}$, for $n \geq \max\{N_1, N_2\}$, for any $\delta &gt; 0$ we can pick a $ y \notin \mathbb{Q}$ such that $| x- y| &lt; \delta$ yet $ | f_n(x) - f_n(y) - 1|  &lt; 2\epsilon$ (by the triangle inequality and using the pointwise convergence properties); so if we choose $\epsilon = 1/10$, say, then we have by continuity that $|f_n(x) - f_n(y)| &lt; 1/10$, yet $| f_n(x) - f_n(y) - 1|  &lt; 1/5$, which (I believe) is absurd, so we have a contradiction.  I'm a bit iffy about whether this is alright or not (I feel I may be subtly abusing the definitions here), so any feedback would be much appreciated.
26884
Does there exist a nowhere monotonic continuous function from some open subset of $\mathbb{R}$ to $\mathbb{R}$? Some nowhere differentiable function sort of object?