# Prove that this integral function is continuous.

• Nov 14th 2009, 11:39 AM
Prove that this integral function is continuous.
If $f \in L^1(m)$ and $F(x) = \int _{- \infty}^{x}f(t)dt$, then F is continuous on $\mathbb {R}$

Proof so far.
Let $x_0 \in X$,

I need to show that $\forall \epsilon > 0$, there exists $\delta > 0$ such that whenever $\mid x - x_0 \mid < \delta$ where $x \in X$, then I will have $\mid F(x)-F(x_0) \mid < \epsilon$

Now, I know have that $\mid F(x)-F(x_0) \mid = \mid \int _{- \infty}^{x}f(t)dt - \int _{- \infty}^{x_0}f(t)dt \mid = \mid \int _{x_0}^{x}f(t)dt \mid$

Since f is integrable, this integral exists, but how should I refine it so it is less than $\epsilon$ ?

Thank you.
• Nov 14th 2009, 11:43 AM
redsoxfan325
$[x_0,x]$ is compact, so $f(t)$ has a max on that interval.

Note: This only works in the reals. I notice you have $X$ as your space, so I'm not sure this applies, unless you know that closed intervals are compact.
• Nov 14th 2009, 04:15 PM
Jose27
Quote:

Originally Posted by redsoxfan325
$[x_0,x]$ is compact, so $f(t)$ has a max on that interval.

How about $f(t)= x^{-1/3}$ if $0<\vert x \vert \leq 1$ and $0$ otherwise. Then $f \in L^1( \mathbb{R} )$ but in any compact set of the form $[-a,b]$ with $a,b \geq 0$ we have that $f$ is not bounded.
• Nov 14th 2009, 04:22 PM
redsoxfan325
Sorry, I was thinking that $f$ was continuous.
• Nov 14th 2009, 05:08 PM
Focus
Quote:

If $\in L^1(m)$ and $F(x) = \int _{- \infty}^{x}f(t)dt$, then F is continuous on $\mathbb {R}$

Proof so far.
Let $x_0 \in X$,

I need to show that $\forall \epsilon > 0$, there exists $\delta > 0$ such that whenever $\mid x - x_0 \mid < \delta$ where $x \in X$, then I will have $\mid F(x)-F(x_0) \mid < \epsilon$

Now, I know have that $\mid F(x)-F(x_0) \mid = \mid \int _{- \infty}^{x}f(t)dt - \int _{- \infty}^{x_0}f(t)dt \mid = \mid \int _{x_0}^{x}f(t)dt \mid$

Since f is integrable, this integral exists, but how should I refine it so it is less than $\epsilon$ ?

Thank you.

Pro tip; $\int_{x_0}^{x}f d\mu=\int_X f \mathbf{1}_{[x_0,x]} d\mu$. What does Fatou say when you look at $x\rightarrow x_0$.
• Nov 14th 2009, 07:22 PM
Fatou's Lemma says that $\int liminf \ f_n \leq lim inf \int f_n$

So as $x_n \rightarrow x_0$, $\int f 1_{ [ x_0,x ] } = \int lim inf \ f 1_{ [x_0,x_n]} \leq lim inf \int f 1_{ [x_0,x_n]} = 0$
• Nov 15th 2009, 05:29 AM
Focus
Quote:

Fatou's Lemma says that $\int liminf \ f_n \leq lim inf \int f_n$

So as $x_n \rightarrow x_0$, $\int f 1_{ [ x_0,x ] } = \int lim inf \ f 1_{ [x_0,x_n]} \leq lim inf \int f 1_{ [x_0,x_n]} = 0$

Sorry I ment to say reverse Fatou.

$\lim\sup \int f\mathbf{1}_{[x_0,x_n]} d\mu \leq \int \lim \sup f\mathbf{1}_{[x_0,x_n]}d\mu=\mu(f \mathbf{1}_{x_0})=0$
for any sequence $x_n \rightarrow x$.

Now if you do apply what that means in epsilon and deltas, you get continuity.
• Nov 15th 2009, 06:24 AM
I have never use the reverse Fatou's lemma. I wonder if this proof can work as well.

Let $\{ x_n \} \rightarrow x_0$.

Then $\lim _{n \rightarrow \infty } \mid F(x_n)-F(x_0) \mid = \lim _{n \rightarrow \infty } \mid \int ^{x_0}_{x_n} f(t)dt \mid$ $= \lim _{n \rightarrow \infty } \int f 1_{[x_n,x_0]}d \mu = \int \lim _{n \rightarrow \infty } f 1_{[x_n,x_0]}d \mu = 0$

Will this be okay as well? Thank you.
• Nov 15th 2009, 02:32 PM
Focus
Quote:

I have never use the reverse Fatou's lemma. I wonder if this proof can work as well.

Let $\{ x_n \} \rightarrow x_0$.

Then $\lim _{n \rightarrow \infty } \mid F(x_n)-F(x_0) \mid = \lim _{n \rightarrow \infty } \mid \int ^{x_0}_{x_n} f(t)dt \mid$ $= \lim _{n \rightarrow \infty } \int f 1_{[x_n,x_0]}d \mu = \int \lim _{n \rightarrow \infty } f 1_{[x_n,x_0]}d \mu = 0$

Will this be okay as well? Thank you.

Well you can't really take the limit inside, you could however use monotone/dominated convergence. If you have not done those tell me what you have done so I can be more helpful.
• Nov 15th 2009, 07:10 PM
But since $1_{[x_n,x_0]}$ is approaching zero, isn't $f 1_{[x_n,x_0]} \leq f$ for all n? Thus I can use the monotone convergence theorem?
But since $1_{[x_n,x_0]}$ is approaching zero, isn't $f 1_{[x_n,x_0]} \leq f$ for all n? Thus I can use the monotone convergence theorem?
Be careful, $\mathbf{1}_{[x_n,x_0]} \rightarrow \delta_{x_0} \neq 0$ where \delta is the Dirac delta function (1 at x_0, 0 everywhere else). Now the Dirac delta function has Lebesgue measure 0 (i.e. integral 0). The sequence $f\mathbf{1}_{[x_n,x_0]}$ is dominated by f (which is integrable), so DMC implies that the integral converges to 0, because DMC tells you that the integral of your sequence converges to $\int f\delta_{x_0}=0$.