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It is well established that sum of a continuous function with a discontinuous function is also discontinuous(See this : Is a continuous function plus a discontinuous function discontinuous?)

However consider this example (NOTE : $[x]$ denotes the greatest integer function.)

$f(x):(0,1){\to}(0,1)$ and $f(x) = x^2$

$g(x):(-\infty,\infty)\to(-\infty,\infty)$ and $g(x)=[x]$ ;

$h(x):(0,1)\to(0,1)$ and $h(x) = f(x) + g(x)$

Clearly, $f$ is continuous, $g$ is discontinuous and $h$ is continuous. So, does this disprove that sum of continuous and discontinuous function is necessarily discontinuous?

An_Elephant
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Yes, some authors implicitly restrict the larger domain to fit the smaller, as you did in the example. In this case the corresponding result is that $h$ is continuous if and only if both restrictions of $f$, $g$ to the domain of $h$ are continuous. As you pointed out, it is crucial that we consider the restrictions.

Formally, take $f_i:\mathcal D_i\rightarrow\mathbb R$ for any $\mathcal D_i\subseteq\mathbb R$. Let $\mathcal D=\mathcal D_1\cap\mathcal D_2$ and $h:\mathcal D\rightarrow\mathbb R$, $x\mapsto f_1(x)+f_2(x)$. Now, we already know that for any $x\in\mathcal D$ the sum $h$ is continuous at $x$ if and only if both $f_1$ and $f_2$ are continuous at $x$. This means that $h$ is continuous (on $\mathcal D$) is continuous if and only if both restrictions of $f_1$, $f_2$ to $\mathcal D$ are continuous.

Matija
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  • Why do different authors use different definitions? Can't there be any agreement and meetings among mathematicians to put a standard ( like there is a IUPAC for chemistry) ? – An_Elephant Oct 02 '22 at 06:50
  • All authors restrict the domain. The sum of two functions $f+g$ is defined as $(f+g)(x)=f(x)+g(x)$. How would you define the sum of two functions where one is defined on a larger domain? – John Douma Oct 02 '22 at 08:14
  • @John Douma: When looking at function spaces, the domain is usually fixed first, say $[0,1]$ for $C([0,1])$, The sum is then introduced as an operator for functions in this space, say $f+g\in C([0,1])$ for $f,g\in C([0,1])$. If you discuss such a given function space and introduce the sum as an operator, you do not (have to) restrict domains. But yes, you may consider $C([0,2])\times C([0,1])\rightarrow C([0,1])$, $(f,g)\mapsto f+g$, or formal sums, and some authors do so (if required). Then, the argument by An_Elephant makes sense. – Matija Oct 02 '22 at 09:27
  • @An_Elephant: Sometimes, different perspectives make sense. Say, matrices (and vectors) may be considered as points or as operators. Measures are encoded as set functions, cumulative distribution functions, Radon-Nikodym derivatives, push-forwards, sometimes as point in a space. All these interpretations are valid. In our case, we may consider the sum $f+g$ of any two functions $f$ and $g$, but only formally, as discussed here, not as an operator. If you want $f+g$ as an operator, you have to restrict the domains, and thus may keep it fixed. – Matija Oct 02 '22 at 09:42
  • @JohnDouma I have been taught to take the common domain for the sum of function. See [![enter image description here][1]][1]

    [1]: https://i.sstatic.net/e4Z72.jpg this . It is on wikipedia .

     – An_Elephant Oct 02 '22 at 11:18