# Measure Theory -- Integration/Continuity

• Oct 27th 2008, 09:19 AM
joeyjoejoe
Measure Theory -- Integration/Continuity
Hey, I am kind of stumped on this one. Anyone want to offer hints? (Cool)

If f is in L1(m) and F(x) = integral ( f(t)dt ) from -infinity to x, then F is continuous on R.

I know a couple things, sure:

**F in L1(m) means that {x : f(x) = infinity} is a null set.
**F is integrable (and therefore continuous) is integral(F(x)) < infinity.
• Oct 27th 2008, 10:47 AM
Laurent
Quote:

Originally Posted by joeyjoejoe
Hey, I am kind of stumped on this one. Anyone want to offer hints? (Cool)

If f is in L1(m) and F(x) = integral ( f(t)dt ) from -infinity to x, then F is continuous on R.

I know a couple things, sure:

**F in L1(m) means that {x : f(x) = infinity} is a null set.
**F is integrable (and therefore continuous) is integral(F(x)) < infinity.

You should read again your definitions: $f\in L^1(m)$ means that $f$ is measurable and integrable with respect to the measure $m$: $\int |f(x)| m(dx)<\infty$. And integrable is not linked with continuous.

To show that the function $F:x\mapsto\int_{-\infty}^x f(t)dt$ is continuous at $x\in\mathbb{R}$, notice that, if $y>x$, $F(y)-F(x)=\int_x^y f(t)dt = \int_{\mathbb{R}} {\rm 1}_{[x,y]}(t)f(t)dt$ (where the 1 is the indicator function). Then you can use the dominated convergence theorem, when $y$ tends to $x$, to show that the previous integral converges to 0.