# Antiderivative is continuous

The following comes from Bass' book on Real Analysis: (Here $dy$ is Lebesgue measure)

Exercise 7.6
Suppose $f:\mathbb{R}\to\mathbb{R}$ is integrable, $a\in \mathbb{R}$, and we define $F(x):=\int_a^x f(y)\, dy$. Show that $F$ is a continuous function.

Another answer on stack exchange covers the case where $f$ is assumed to be Riemann integrable, with $f$ and $F$ being functions from some $[a,b]$. The proof there uses the fact that $f$ is Riemann integrable if and only if it is bounded and continuous almost everywhere.

Folland's book on analysis appears to define integration for functions $f:X\to[-\infty,\infty]$ on the extended real numbers. ("The integral defined in the previous section [for functions $f:\mathbb{R}\to[0,\infty]$] can be extended to real-valued measurable functions $f$..." Except, real-valued means $\mathbb{R}$ but extending the old definition forces the inclusion of infinities. However, as far as I can tell the actual definition as integrating the positive and negative parts agrees with taking $[-\infty,\infty]$-valued functions.

I only bring this up because it seems exercise 7.6 should hold for: $f:\mathbb{R}\to[-\infty,\infty]$. $$f(x):=\begin{cases} x^{-1/2}& x>0\\ (-x)^{-1/2} &x<0 \\ \infty & x=0\end{cases}$$

In trying a few things I also wondered if the $F$ from 7.6 is always measurable. (this should be an easy yes?)

It's worth noting that chapter 7 of Bass covers the usual monotone, dominated, etc. theorems.

The question then is if the modified exercise is correct, how different are the lebesgue measure -vs- riemann measure proofs, how to go about either version (Lebesgue preferred for the modified exercise)

• Is $\displaystyle \int f \, \mathrm{d} y$ defined as $\displaystyle \int f^+ \, \mathrm{d} y + \displaystyle \int f^- \, \mathrm{d} y$ and $\displaystyle \int f^+ \, \mathrm{d} y = \sup \left\{ \int s \, \mathrm{d} y : s \leqslant f \text{ is a simple function, i.e. has finite range} \right\}$ ? May 20, 2015 at 6:33
• Yes, that's what I'm using :)
– ttt
May 20, 2015 at 6:35
• I could ask the same question except defining the function to be zero at zero. Though, the function is only infinity on a null set so it shouldn't affect F. (I'm saying modifying f on a null set doesn't change F)
– ttt
May 20, 2015 at 6:53

I believe the proof is somewhat different than the case of Riemann integration. Also, as a comment noted, allowing functions to take values $\pm \infty$ is harmless (an integrable function can only take these values on a measure zero set). All integrable functions are measurable; this may be easy or may take some work, depending on exact definitions.

One proof of the main exercise uses the dominated convergence theorem. Fix $x_0$. For convenience, let $\chi_{(x_0,x)}$ denote the characteristic function of the interval when $x_0 < x$, and denote $-\chi_{(x,x_0)}$ otherwise. Then $$\lim_{x \to x_0} F(x) - F(x_0) = \lim_{x \to x_0} \int_{x_0}^x f(y)dy = \lim_{x\to x_0} \int_{-\infty}^{\infty} \chi_{(x_0,x)}f(y)dy = \int_{-\infty}^{\infty}\lim_{x\to x_0} \chi_{(x_0,x)}f(y)dy = \int_{-\infty}^{\infty} 0 dy = 0\text{.}$$

• As far as I know, the implication $\text{integrable} \implies \text{measurable}$ is quite lengthy to show with all detail. May 20, 2015 at 7:36
• @Adayah, you're right, it's not totally immediate using the definition given in your comment to the OP. It's easy for some others, though. I edited my post.
– user231101
May 20, 2015 at 7:43
• I was asking if $F$ (not $f$) was measurable provided $f$ was integrable. But this is also good to know.
– ttt
May 20, 2015 at 15:42
• So just to be clear: When you write $\int_{x_0}^x$, as far as Lebesgue integrals go thats code for $\int_\mathbb{R} \chi_{(x_0,x)}$ or $-\int_\mathbb{R}\chi_{(x,x_0)}$ depending on if $x_0<x$ or the other way around. Since its just a +/- you can let the notation "handle" the cases.
– ttt
May 20, 2015 at 16:43
• @ttt Exactly. The point is that it should be $F(x) - F(x_0)$ in either case.
– user231101
May 20, 2015 at 20:22

Since

$$\int f \, \mathrm{d} y = \int f^+ \, \mathrm{d} y + \int f^- \, \mathrm{d} y,$$

it suffices to show that

$$G(x) = \int \limits_a^x f^+ \, \mathrm{d} y$$

is continuous and the other part will follow analogously. Fix $x_0 \in \mathbb{R}$ and $\varepsilon > 0$. We will search for such $\delta > 0$ that $|G(x_0) - G(x)| < \varepsilon$ i.e. $\displaystyle \left| \, \int \limits_x^{x_0} f^+ \, \mathrm{d} y \, \right| < 3\varepsilon$ whenever $|x - x_0| < \delta$.

Since by definition

$$\int \limits_{x_0-1}^{x_0+1} f^+ \, \mathrm{d} y = \sup \left\{ \int \limits_{x_0-1}^{x_0+1} s \, \mathrm{d} y : s \leqslant f \text{ is a simple function} \right\},$$

there is such partition $(E_i)_{i=1}^n$ of $[x_0-1, x_0+1]$ and a simple nonnegative function $\displaystyle s(x) = \sum_{i=1}^n v_i \cdot \mathbf{1}_{E_i} \leqslant f(x)$ that

$$\sum_{i=1}^n v_i \cdot \lambda(E_i) \leqslant \int \limits_{x_0-1}^{x_0+1} f^+ \, \mathrm{d} y < \varepsilon + \sum_{i=1}^n v_i \cdot \lambda(E_i).$$

Let $M = \max\{ v_i : i = 1, 2, \ldots, n \}$ and $\delta = \min \{ \frac{\varepsilon}{M}, 1 \}$. For $|x - x_0| < \delta$ clearly

$$\left| \, \int \limits_x^{x_0} f^+ \, \mathrm{d} y \, \right| \leqslant \int \limits_{x_0-\delta}^{x_0+\delta} f^+ \, \mathrm{d} y \qquad \text{ and } \qquad \int \limits_{x_0-\delta}^{x_0+\delta} ( f^+ - s) \, \mathrm{d} y \leqslant \int \limits_{x_0-1}^{x_0+1} ( f^+ - s) \, \mathrm{d} y < \varepsilon$$

so

\begin{align} \int \limits_{x_0-\delta}^{x_0+\delta} f^+ \, \mathrm{d} y & < \varepsilon + \int \limits_{x_0-\delta}^{x_0+\delta} s \, \mathrm{d} y = \varepsilon + \sum_{i=1}^n v_i \cdot \lambda( E_i \cap [x_0-\delta, x_0+\delta]) \\[1ex] & \leqslant \varepsilon + \sum_{i=1}^n M \cdot \lambda( E_i \cap [x_0-\delta, x_0+\delta]) \\[1ex] & = \varepsilon + M \cdot \lambda( [x_0-\delta, x_0+\delta] ) = \varepsilon + 2 M \delta \leqslant 3 \varepsilon \end{align}

• Thanks for this additional approach! Cool.
– ttt
May 20, 2015 at 21:05