# Thread: Prove that this integral function is continuous.

1. ## Prove that this integral function is continuous.

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

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

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

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

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

Thank you.

2. $\displaystyle [x_0,x]$ is compact, so $\displaystyle f(t)$ has a max on that interval.

Note: This only works in the reals. I notice you have $\displaystyle X$ as your space, so I'm not sure this applies, unless you know that closed intervals are compact.

3. Originally Posted by redsoxfan325
$\displaystyle [x_0,x]$ is compact, so $\displaystyle f(t)$ has a max on that interval.
How about $\displaystyle f(t)= x^{-1/3}$ if $\displaystyle 0<\vert x \vert \leq 1$ and $\displaystyle 0$ otherwise. Then $\displaystyle f \in L^1( \mathbb{R} )$ but in any compact set of the form $\displaystyle [-a,b]$ with $\displaystyle a,b \geq 0$ we have that $\displaystyle f$ is not bounded.

4. Sorry, I was thinking that $\displaystyle f$ was continuous.

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

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

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

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

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

Thank you.
Pro tip; $\displaystyle \int_{x_0}^{x}f d\mu=\int_X f \mathbf{1}_{[x_0,x]} d\mu$. What does Fatou say when you look at $\displaystyle x\rightarrow x_0$.

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

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

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

So as $\displaystyle x_n \rightarrow x_0$, $\displaystyle \int f 1_{ [ x_0,x ] } = \int lim inf \ f 1_{ [x_0,x_n]} \leq lim inf \int f 1_{ [x_0,x_n]} = 0$
Sorry I ment to say reverse Fatou.

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

Now if you do apply what that means in epsilon and deltas, you get continuity.

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

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

Then $\displaystyle \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 $$\displaystyle = \lim _{n \rightarrow \infty } \int f 1_{[x_n,x_0]}d \mu = \int \lim _{n \rightarrow \infty } f 1_{[x_n,x_0]}d \mu = 0 Will this be okay as well? Thank you. 9. Originally Posted by tttcomrader I have never use the reverse Fatou's lemma. I wonder if this proof can work as well. Let \displaystyle \{ x_n \} \rightarrow x_0 . Then \displaystyle \lim _{n \rightarrow \infty } \mid F(x_n)-F(x_0) \mid = \lim _{n \rightarrow \infty } \mid \int ^{x_0}_{x_n} f(t)dt \mid$$\displaystyle = \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.

10. Yes, I have done both Monotone and DCT.

But since $\displaystyle 1_{[x_n,x_0]}$ is approaching zero, isn't $\displaystyle f 1_{[x_n,x_0]} \leq f$ for all n? Thus I can use the monotone convergence theorem?

But since $\displaystyle 1_{[x_n,x_0]}$ is approaching zero, isn't $\displaystyle f 1_{[x_n,x_0]} \leq f$ for all n? Thus I can use the monotone convergence theorem?
Be careful, $\displaystyle \mathbf{1}_{[x_n,x_0]} \rightarrow \delta_{x_0} \neq 0$ 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 $\displaystyle f\mathbf{1}_{[x_n,x_0]}$ 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 $\displaystyle \int f\delta_{x_0}=0$.