# Why is $\mathfrak{L}^1$ not a vector space of functions that take infinity as a value

In measure and integration theory we sometimes work with measurable functions $$f: (X, \mathfrak{A}) \to (\overline{\mathbb{R}}, \overline{\mathfrak{B}}),$$ where $(X,\mathfrak{A})$ denotes an arbitrary measure space, $\overline{\mathbb{R}}:= \mathbb{R}\cup \{\infty,-\infty\}$ and $\overline{\mathfrak{B}}$ the $\sigma$-algebra of the Borel sets. In addition to the "natural" definitions of sum and multiplication in this space the author of my book (Elstrodt; german reference) "arbitrarily" defines $$0 \cdot (\pm \infty) := (\pm \infty) \cdot 0 =:= 0,\quad \infty - \infty := -\infty + \infty := 0.$$

• Later on, there appears a theorem stating that for two measurable functions $f,g$ defined as above and $\alpha, \beta \in \overline{\mathbb{R}}$ their linear combination $\alpha f + \beta g$ is again measurable.
• In the chapter on Lebesgue-integrable functions, there is a theorem for $f, g$ as above with measure $\mu$ on $\mathfrak{A}$ and $\alpha, \beta \in \mathbb{R}$ (no overline intended!), where he proves that $\alpha f + \beta g$ is again Lebesgue-integrable with $$\int_X (\alpha f + \beta g) \mathrm{d}\mu = \alpha \int_X f \mathrm{d}\mu + \beta \int_X g \mathrm{d}\mu.$$

Finally, he remarks that the set of the Lebesgue-integrable functions with values in $\overline{\mathbb{R}}$ does not form a vectorspace with respect to pointwise addition, if there is a non-empty set of $\mu$-measure 0.

Now my questions:

• When we talk about vector spaces, in this context we mean vector spaces over the field $\mathbb{R}$, since $\overline{\mathbb{R}}$ is not a field, right?
• Do the measurable functions form a vector space over $\mathbb{R}$?
• Do you understand the author's remark? Can you give an example?
• Many other authors define the vector space of Lebesgue-integrable functions on functions which only take values in $\mathbb{R}$, since "$\infty - \infty$ is not defined". I think this is a poor explanation, since we can indeed define this difference and it won't affect the integrability of the difference of two functions since these values are only attained on sets of $\mu$-measure zero. What am I missing?
• What about $f = \frac1x$ and $g = -\frac1x + 1$? Then $f+g = 1$ but $f(0) + g(0) \ne 1$. – AlexR Aug 9 '15 at 19:15
• @Alex Is that the case that $f+g=1$? How are you defining $f$ and $g$ at $0$?. He didn't say $1/0:= \infty$ at any point. – Aloizio Macedo Aug 9 '15 at 19:17
• @AloizioMacedo As long as you agree that $f-f = 0$, you'll have to also agree that $f+g = f-f+1 = 1$. But $f-f = 0$ is one of the corollaries of the vector space axioms. – AlexR Aug 9 '15 at 19:19
• I've never seen the definition $\infty - \infty =0$ before. – zhw. Aug 9 '15 at 19:21
• @zhw I haven't either.. Authors usually avoid trying to define that operation since it leads to a lot of contradictions if you want to keep associativity. I feel like we are missing something important here.. Perhaps this is an exercise to show that defining a difference of infinities is a bad idea. – Cameron Williams Aug 9 '15 at 19:25

Indeed, usually one allows non-negative functions to take the value $+\infty$, but one usually doesn't talk about integrable functions taking values in $\overline{\Bbb R}$. There are good reasons for that; I shuddered when I saw that definition $\infty-\infty=0$. I hope his main point in doing this is to point out that it's not a good idea...
Anyway. If the empty set is the only null set then an integrable function is finite everywhere and so the integrable functions form a vector space. (A vector space over $\Bbb R$; no, $\overline{\Bbb R}$ is not a field.)
But now say $E\ne\emptyset$ and $\mu(E)=0$. Now an integrable function can take infinite values, and we no longer have a vector space. One of the axioms for a vector space is this: $$(c+d)f=cf+df$$for any vector $f$ and scalars $c,d$. Say there's a point $x$ where $f(x)=+\infty$. Let $c=2$ and $d=-1$. Then $$(cf(x))+(df(x))=(2(\infty))-\infty=\infty-\infty=0,$$while $$(c+d)f(x)=(1)\infty=\infty.$$ So $(c+d)f\ne cf+df$. Bad.
And there's no way to define $\infty-\infty$ to fix this.