# Indefinite Integral $F(x)=\int_{-\infty}^xf(t)dm(t)$ Proof Check

I have the following proposition and proof, and I was just wondering if anyone could tell me if I was correct in my reasoning or not.

Let $f$ be a non negative integrable function. Show that the function $F$ defined by $F(x)=\int_{-\infty}^xf(t)dm(t)$ is continuous, where $m$ denotes the Lebesgue measure.

Proof: Since $f$ is integrable, we have that $\int_{-\infty}^{\infty}f(t)dm(t)=K$ for some $K \in \mathbb{R}$, and this gives us that $lim_{x\rightarrow \infty} \int_{-\infty}^xf(t)dm(t)=K$. From this, given $\epsilon > 0$, there exists an $N$ such that for all $x_1,x_2 \geq N$, we have that $K-\int_{-\infty}^{x_1}f(t)dm(t)<\frac{\epsilon}{2}$ and $K-\int_{-\infty}^{x_2}f(t)dm(t)<\frac{\epsilon}{2}$. So we have that $|F(x_1)-F(x_2)|=|\int_{-\infty}^{x_1}f(t)dm(t)-\int_{-\infty}^{x_2}f(t)dm(t)|$, and this is $\leq |K-\int_{-\infty}^{x_1}f(t)dm(t)|+$ $|K-\int_{-\infty}^{x_2}f(t)dm(t)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

I feel like all the components of the proof are there but it still feels a bit dodgy to me. What am I missing, and where are my errors in logic?

• This is really wrong. You're not showing that the function is continuous at (say) $x=-1$. Also, I believe that in your question, it should be $F(x)$ not $F(t)$, since $t$ isn't an actual variable in your formula. Jan 9, 2013 at 0:47
• So how would you go about it? Jan 9, 2013 at 0:50
• How would you show that $g(x)$ is a continuous function for all $x \in (-\infty, \infty)$? Replace $g$ with $F$. Jan 9, 2013 at 0:51
• By picking a $\delta$ based on $\epsilon$ and $x$. I'm having problems finding a $\delta$ that works. Jan 9, 2013 at 0:56

Fix $x_0$. We have:
$$|F(x) - F(x_0)| \le \int_{x_0}^x |f(t)| \, dm(t)$$
Since $f$ is integrable, $|f|dm$ is bounded and absolutely continuous with respect to $m$. Hence, for each $\varepsilon > 0$, we can find $\delta > 0$ so that:
$$|x - x_0| < \delta \Rightarrow \int_{x_0}^x |f(t)| \, dm(t) < \varepsilon$$
• How does $f$ integrable imply bounded? What about $f(x)=\frac{1}{\sqrt{x}}$? Jan 9, 2013 at 1:05
• The measure defined by $|f|dm$ is bounded, not the function itself. Jan 9, 2013 at 1:06