When two functions are equal, but not.

I haven't looked into it much, but this is something I've been aware of that I know I need to look into.

When I have a function $f(x)=\frac{x+1}{x+1}$, There is a discontinuity at $x=-1$, yet $\frac{x+1}{x+1}=1$ and has no discontinuity. It's like they're equal but not.

The qualities of the function are not preserved after the algebraic manipulation, so I can't strictly say that $\frac{x+1}{x+1}=1$.

This is an issue for me when understanding integrals. For instance, finding the definite integral of the quotient, if the discontinuity is within my limits, doesn't make sense. But after changing the quotient to a constant, it's possible: but I've found the area under a curve that wasn't complete. I've found a solution for an unanswerable, insensible question.

I hope I've made this clear. My question is, is this right? How do I come to terms with this?

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There are a few ways to look at it. First, certainly $\frac{x + 1}{x+1}$ is a way of writing $1$. However, if you insist that the function so defined should be defined everywhere but $-1$, then it is the same as defining a function $f$ on $(-\infty, -1)\cup (-1, \infty)$ by $f(x) = 1$. This function extends to $x = -1$ continuously. – William Jan 8 '12 at 18:10
This question made me think: for continuity purposes, $\frac{x+1}{x+1} \ne 1$ in Leibniz sense. However, if one defines equivalency classes of continuous functions, then $\frac{x+1}{x+1} \equiv g(x),$ where $g(x)= \begin{cases} 1,&\quad x\neq -1 \\ \mbox{undefined},&\quad x= -1 \end{cases}.$ – user2468 Jan 8 '12 at 18:37
Informally, the discontinuity is infinitely thin, and thus has no bearing on the area under the curve; formally, you use an improper integral to avoid doing anything nonsensical. – Jon Purdy Jan 9 '12 at 5:03

A function is an object with three parts: a set of inputs called its domain, a set of allowable outputs called its range, and a rule tying each element of it domain to its range.

If you fail to specify all three of these things, you give an incomplete specification of the function.

So when you say $$f(x) = {x + 1\over x + 1},$$ you are omitting the intended domain. The natural domain is the set of all real numbers save for $-1$. This is the tacit domain often used in discussing real functions of a a real variable.

The constant function $x\mapsto 1$ defined on the whole real line is an extension of $f$ to the entire line. This is a different function, since its domain is larger than that of $f$.

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$x=-1$ is what's called a removable singularity of your function $f$: if you redefine your function at that point it becomes continuous.

As for integration, the definition of the Riemann integral does require the function to be defined everywhere in the interval in question. However, for a function which is undefined at some point you could consider an improper integral, which would converge to the correct value. When you get to Lebesgue integration, having the function undefined at some point is allowed, and won't make a difference.

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${x+1\over x+1}=1$ only when $x\ne -1$. Your function $f$ is not defined at $x=-1$.
The graph of $f$ has a "hole" in it:

Things can be made nice, though: You could redefine $f$ to take the value $1$ at $x=-1$. This would give you a new function $\tilde f$ that differs from $f$ at only one point. Also, $\tilde f$ is identically 1; and so, it can be integrated over any interval of finite length.

Strictly speaking the definite integral of $f$ over $[-2,1]$ is undefined; but, you should be able to convince your self that the area under the graph of $f$ over $[-2,1]$, say, is 3, since the area under the "hole" is 0. Better yet, imagine a rectangle of height 1 and very small width centered at $x=-1$. The area of that rectangle is very small; so the area under the graph of $f$ over $[-2,1]$ is 3 (after taking appropriate limits). This is what Paccio is doing in his answer.

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Your last paragraph is glossing over the concept of impulse functions e.g. Dirac delta. – wim Jan 8 '12 at 23:20
@wim: There are many people (myself included) who do not consider the delta function to be a function. (Though perhaps you're right that it still needs mentioning, even if just to say that.) – Billy Jan 9 '12 at 14:47

If you look at functions from a set-theory point of view, they are nothing but triplets $f = (E,F,\Gamma)$, $E$ and $F$ being the domain and the codomain, $\Gamma$ being a subset of $E \times F$ — the graph — such that for all $x \in E$, there is an unique $y \in F$ such that $(x,y) \in \Gamma$. We then note $f(x) = y$.

We then say that two functions $(E,F,\Gamma)$ and $(E',F',\Gamma')$ are equals if $E=E'$, $F=F'$ and $\Gamma=\Gamma'$.

When one defines a function by a symbolic expression, such as $f(x) = \frac{x+1}{x+1}$, it is only a shorthand to define $f= (E,F,\Gamma)$, it is implied that

• $E$ is the biggest subset of $\mathbb{R}$ such that for all $x \in E$, $\frac{x+1}{x+1}$ is defined
• $F$ is the smallest subset of $\mathbb{R}$ such that for all $x \in E$, $\frac{x+1}{x+1} \in F$
• $\Gamma = \{(x,y) \in E \times F ~|~ y = \frac{x+1}{x+1}\}$

Since $\frac{0}{0}$ is undefined, $f$ would not be defined in $0$, but since this aimed to be a shorthand, we usually extends the $E$ (and subsequently $F$ and $\Gamma$) to the points $x \in \bar{E}$ — the closure of $E$ — such that $f$ has a limit $y$ in $x$. We then note $f(x)=y$, so formally this defines an extension of $f$, that we also note $f$, which is not equals to the first $f$, since they have neither the same domain, nor the same graph.

Long story short, the function defined by $f(x) = \frac{x+1}{x+1}$ is equal to $1$ and defined in $\mathbb{R}\smallsetminus \{-1\}$ or in $\mathbb{R}$, depending on the context.

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You should probably think of $\frac{x+1}{x+1}$ as having a singularity - or at least some kind of blip - at x = -1. This is the honest answer - we cannot get away from the fact that crudely plugging x = -1 into that expression gives 0/0. But sort of "morally" it's not too hard to skip over this singularity, for the obvious reasons: it is a very small singularity (a single point) and we can fill it in in a natural way that makes the function continuous there. A singularity of this form is called "removable", and we tend to identify the two functions even if this isn't strictly true. There are other types of singularities that are harder (or impossible) to remove: poles and essential singularities, for example. You might want to look these up - they will be mentioned in detail in a book on complex analysis.

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"we tend to identify the two functions even if this isn't strictly true" - Yes, indeed, because it is very convenient to do so, and it would be helpful to beginners to point out that such convenience-identification starts even earlier, for example, using "contains" to mean "is a superset of", and considering a 3-tuple to be either (a,(b,c)) or synonymously ((a,b),c), or even a sequence of length 3. Furthermore, I daresay, as the theory progresses, such convenience-identifications occur more and more frequently. For example, homeomorphic topological spaces are identified, are they not? – Hexagon Tiling Jan 9 '12 at 9:30
@HexagonTiling: I'm not sure. In my mind, a 3-tuple is a very fixed and well-defined object. You may define it differently if you like, and indeed if you're working very strictly inside some logical system maybe your hand is forced here, but it is only when we collaborate that we make the implicit identification between our two definitions. Homeomorphic topological spaces are most definitely not identified except when we have decided that we only care about the topology - elliptic curves are a good example of this. Though perhaps that's important to point out in its own right. :) – Billy Jan 9 '12 at 14:54

The function $f(x)=\frac{x+1}{x+1}$ is not well-defined at $x=-1$. If you define $$f(x)= \begin{cases} \frac{x+1}{x+1},&\quad x\neq -1 \\ 1,&\quad x= -1 \end{cases}$$ then $f$ is really equal to the constant function $1$.

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Assume your $f(x)$ is discontinuous in $x_0$ but continuously extends to the whole real line; then let $\bar{f}(x)$ its unique extension.

The integral of $f(x)$ and of $\bar{f}(x)$ coincide whether the limits of integration are: in fact, let $a < x_0 < b$ (which is the only problematic case) and write: $$\begin{split}\int_a^b f(x)\ \text{d} x &=\int_a^{x_0} f(x)\ \text{d} x +\int_{x_0}^b f(x)\ \text{d} x \\ &=\lim_{\varepsilon \to 0^+}\int_a^{x_0-\varepsilon} f(x)\ \text{d} x +\lim_{\delta \to 0^+}\int_{x_0+\delta}^b f(x)\ \text{d} x\\ &=\lim_{\varepsilon \to 0^+}\int_a^{x_0-\varepsilon} \bar{f}(x)\ \text{d} x +\lim_{\delta \to 0^+}\int_{x_0+\delta}^b \bar{f}(x)\ \text{d} x\\ &= \int_a^{x_0} \bar{f}(x)\ \text{d} x +\int_{x_0}^b \bar{f}(x)\ \text{d} x \\ &= \int_a^b \bar{f}(x)\ \text{d} x\; .\end{split}$$ Remark that you can replace $f(x)$ with $\bar{f} (x)$ into $\int_a^{x_0-\varepsilon}$ and $\int_{x_0+\delta}^b$ because your are performing integration far away from the discontinuity point $x_0$.

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I am a bit puzzled at this. Could you please clarify what you are trying to demonstrate in the context of the original question? For example, we can consider two Lebesgue integrable functions which coincide outside a set of measure zero; but this set can be dense. Are you trying to show that in some "category" they are the same? Sorry, I am a bit puzzled. – William Jan 8 '12 at 18:25
Well, I`m trying to prove that if a function $f$ is continuous in $\mathbb{R}\setminus \{ x_0\}$ and if it has a removable discontinuity in $x_0$ (just as in the original question) then, for all $a<x_0<b$, the improper integral $\int_a^b f(x)\ \text{d} x := \lim_{\varepsilon \to 0^+} \int_a^{x_0-\varepsilon} f(x)\ \text{d} x +\lim_{\delta \to 0^+} \int_{x_0+\delta}^b f(x)\ \text{d} x$ equals the Riemann integral $\int_a^b \bar{f}(x)\ \text{d} x$, where $\bar{f}$ is the continuous extension of $f$ to the whole real line... – Pacciu Jan 9 '12 at 16:52

There is no discontinuity at x=−1, just take the right and left limits and you'll see that. Hence f = 1.

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The function is not defined since $-1+1=0$ and you cannot write $0$ in the denominator, however the discontinuity is of a removable type since the limits from both sides are different. – Asaf Karagila Jan 8 '12 at 23:43
@Asaf Karagila: "are different"? I think you mean "are the same". – Hexagon Tiling Jan 9 '12 at 9:16
@Hexagon: Of course, thank you. – Asaf Karagila Jan 9 '12 at 9:26
does the lim_{y->0}{ y / y } not go to 1? – j13r Jan 16 '12 at 19:38