# Calculus, integrable function

Let $f:[a,b]\to \mathbb{R}$ be integrable. Show that the following statements are equivalent:

1. $\int_a^b \lvert f(x) \rvert dx = 0$

2. If $f$ is continuous at $c$, then $f(c)=0$

3. The interior of $X=\{x\in[a,b]; f(x)\neq 0\}$ is empty, i.e. $\operatorname{int}(X) = \emptyset$

I proved $(1) \implies (2)$, but I am struggling to prove the remaining cases.

Case $(1)\implies (2)$: Suppose $f$ is continuous at $c$ and $f(c)\neq 0$, then $\lvert f(c) \rvert > 0$ where $\lvert f \rvert$ is also continuous at $c$. Then there is an open $I$ such that $x\in I$ then $\lvert f(x) \rvert > 0$. Let $[c_1,c_2]\subset I$, then

$$\int_a^b \lvert f(x) \rvert dx = \int_a^{c_1} \lvert f(x) \rvert dx + \int_{c_1}^{c_2} \lvert f(x) \rvert dx + \int_{c_2}^b \lvert f(x) \rvert dx > 0$$

Contradiction. Then $f(c)=0$.

• Do you know that f Riemann integrable implies f is continuous a.e.? – zhw. Aug 13 '16 at 18:54
• No, I didn't know that. I am just beginning the chapter on Riemann integrals. The definition so far of "integrable" is that the lower and upper sums converge to the same quantity. – Michael Aug 13 '16 at 18:56

For 3) implies 1): The first thing to note is that since $f$ is Riemann integrable on $[a,b],$ so is $|f|.$ For any partition $P$ of $[a,b],$ each subinterval must contain a point not in $X.$ (Otherwise $X$ contains a subinterval.) So $f$ vanishes somewhere in each subinterval. It follows that for every partition, the lower sum for $|f|$ is $0.$ Since $\int_a^b|f|$ is the supremum of all lower sums, we have $\int_a^b|f|=0.$
For $(2)\Rightarrow(3)$: Let's first show that the continuity points of any Riemann integrable function is dense in $[a,b]$. Suppose this is not the case.
Define the oscillation of $f$ at a point $x$ as $\omega_f(x)=\inf\left\{\operatorname{diam}f(U):U\text{ open}, x\in U\right\}$, and $O_\epsilon=\left\{x\in[c,d]:\omega_f(x)<\epsilon\right\}$. Note that $O_\epsilon$ is open, and that the set of continuity points of $f$ (in $[c,d]$) is $\bigcap_mO_{1/m}$, which we are assuming to be non-dense. By Baire's theorem, at least one of the $O_{1/m}$ is non-dense in $[a,b]$, so there is an interval $[e,f]$ which does not intersect $O_{1/m}$.
Now we note the following: if $y$ is any point in $[e,f]$, and $I$ is any interval containing $y$ in its interior, $\operatorname{diam}f(I)\geq 1/m$, because $\omega_f(y)\geq 1/m$ and by the definition of oscillation.
Let $P$ be a partition of $[a,b]$ which is fine enough, in the sense that $[c,d]$ is a union of elements of $P$. Name these elements $I_1,\ldots,I_n$. Each $I_i$ contains a point of $[e,f]$ in its interior, so the difference of the upper and lower Riemann sums for $P$ is $$\sum_{I\in P}\operatorname{diam}(f(I))\cdot\operatorname{length}(I)\geq\sum_{i=1}^n\frac{1}{m}\operatorname{length}(I_i)=\frac{f-e}{m}$$ so the difference between upper and lower Riemann sums is bounded below, contradicting Riemann integrability of $f$. We conclude that the set of continuity points of $f$ is dense.
Now suppose the interior of $X$ was nonempty. Then $X$ would contain a continuity point $x$, which would satisfy $f(x)\neq 0$, contradicting (2).