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?

 A: 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.
