# Does the meaning of “continuous function” became “almost everywhere continuous function” in $L^1$ space?

Let $L^1$ be the space of Lebesgue integrable functions, to make $$\|f\|_1=\int|f|$$ a norm we need to indentify functions that are equal almost everywhere, which means that the elements of $L^1$ are not functions but classes of equivalence of functions.
So, does the the proposition

The subspace of continuous function is not closed (as a subspace of $L^1$)

mean there are sequences of functions continuous a.e. whose limit is discontinuous in a set of positive measure?

Is that the meaning of the proposition above?

• Strictly speaking, the set of continuous functions isn't even a SUBSET of $L^1$. – user223391 Jan 12 '18 at 17:35
• @ZacharySelk well I guess it is meant "the subspace of integrable continuous functions" – la flaca Jan 12 '18 at 17:40
• Strictly speaking, the set of integrable continuous functions is not even a SUBSET either. – user223391 Jan 12 '18 at 17:52
• $L^1$ does not contain any functions, strictly speaking. – user223391 Jan 12 '18 at 17:54
• @ZacharySelk ah ok, I guess you are refering to the abuse of language. Yes, I agree, that's why that proposition seems a little bit ambiguous to me – la flaca Jan 12 '18 at 17:55

The point is not that the limit is discontinuous on a set of positive measure. In fact that does not even really make sense, because $L^1$ limits and a.e. limits are not unique at the level of single functions, so "the limit" in this sense doesn't mean anything. Instead, the point is that there exists a sequence of classes, each with a continuous representative, whose $L^1$ (or a.e.) limit has no continuous representative.
In general a class containing a continuous function is considered to be "canonically represented" by that function. For instance the zero vector in $L^1(\mathbb{R})$ is technically represented by $1_{\mathbb{Q}}$ but we prefer to represent it by the zero function.
• @laflaca Just in general "the" a.e. or $L^1$ limit of a sequence of functions (not classes) is still a whole equivalence class, not one function. This is the usual situation for pseudometric spaces. – Ian Jan 12 '18 at 17:54
Being continuous means that there is an element inside the class that is continuous everywhere (which implies that many elements in the class is continuous almost everywhere, but saying that some elements inside a class are continuous almost everywhere is strictly weaker : take for example the characteristic function of an interval, every element in the class is continuous almost everywhere, but none of them is continuous everywhere, so it's not continuous in $L^1$).
• If I understand your intent, your parenthetical is wrong. Compare $0$ and $1_{\mathbb{Q}}$. – Ian Jan 12 '18 at 17:36