# question concerning continuous and locally bound functions in relation to continuous almost everywhere

I have a question from Munkres' Analysis on Manifolds text, under the section 15 Improper Integrals, question 8. $\newcommand{\R}{\mathbb{R}}$

The question reads as follows: Let $A$ be open in $\R^n$. We say $f: A \to \R$ is locally bounded on $A$ if each $x \in A$ has a neighborhood on which $f$ is bounded. Let $F(A)$ be the set of all functions $f:A \to \R$ that are locally bounded on $A$ and continuous on $A$ except on a set of measure zero.

(a) Show that if $f$ is continuous on $A$, then $f$ is an element of $F(A)$.

(b) Show that if $f$ is in $F(A)$, then $f$ is bounded on each compact subset of $A$ and the definition of the extended integral, the integral of $f$ over $A$, goes through without change.

Before I can do (b), I am having trouble with (a). But I thought a solution might go something like this.

Suppose $f$ is not an element of $F(A)$, meaning suppose the set of discontinuities of $f$ over $A$ is not of measure zero. Since the set of discontinuities of $f$ over $A$ is not of measure zero, $f$ is not integrable on the entirety of $A$. However, $f$ is continuous on $A$, so for any element $x \in A$, and a M, we can find an open neighborhood $V_{x}$ such that $|f(x)| \le M$, and $f(V_{x})$ is a subset of $f(A)$. (Here continuity would easily force $f$ to be locally bounded). $f$ is both locally bounded and continuous on $A$, $f$ is integrable over $V_{x}$. Since it was assumed that $f$ is not integrable on any portion over $A$. Hence we have a contradiction.

I think the point of this question is related to how bad discontinuities can get in the situation of improper integrals. I know the classification of discontinuities have to do with Baire category theorems. But I am not allowed to use that even if I know that part of analysis in depth.

Also, in Munkres' text, seqyebce if compact sets were introduced in improper integral section, while partition of unity is introduced in the chapters on change of variables which comes after.

My apologies in advanced if my post seem very not LaTex like. I don't use LaTex enough and each time, I even forget the simplest syntax.

Clearly f is continuous on A except on a set of measure 0 since it is a continuous function on A. Since f is continuous then for x in A and in a small delta ball (chosen small enough that it is contained in A) around b in A we have $\vert f(x)-f(b)\vert<\epsilon$. So $\vert f(x)\vert <\vert f(b)\vert +\epsilon$ for all x in the delta ball. x was arbitrary so f is locally bounded. Hence, $f\in F(A)$.
• So you want to prove $f\in F(A)$ when $A$ is an open set? For each $x\in A$ we have a neighbourhood where $f$ is bounded. Since $A$ is open then we have a neighbourhood of $x\in A$ that is completely contained in $A$. By taking the intersection of the two neighbourhoods we can avoid the boundary of $A$ and have $f$ bounded on the neighbourhood. So to prove local boundedness when $A$ is open we do not need to worry of boundary points since small neighourhoods are disjoint from the boundary of $A$. – user71352 Aug 16 '13 at 20:45
• It is possible for a function to be continuous on an open set but have the set of discontinuities at the boundary not of measure $0$. For example take the indicator of the open removed sets of a fat cantor set. The discontinuities are all on the boundary but the boundary does not have measure $0$. I don't see why $f$ being discontinuous on the boundary of $A$, when $A$ is open, affects whether $f$ is a member of $F(A)$. We only care about discontinuities of $f$ that lie in $A$ and continuity is determined locally. – user71352 Aug 16 '13 at 20:56