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Let $f:[0,1]rightarrowmathbb{R}$ be a continuous function.

1. Show that for each $epsilonin(0,1)$, $limlimits_{nrightarrowinfty}intlimits_0^{1-epsilon}f(x^n)dx=(1-epsilon)f(0)$
   


2. Find $limlimits_{nrightarrowinfty}intlimits_{0}^{1}f(x^n)dx$.
   
   Hint: Start by explaining why $f$ is bounded.

For my Answer:,

I have done the part one by using the fact that
begin{align*}
left|intlimits_{0}^{1-epsilon}f(x^n)dx-(1-epsilon)f(0)right|leq intlimits_0^{1-epsilon}|f(x^n)-f(0)|dx
end{align*}
And with the continuity of $f$ at zero along with $x^nleq(1-epsilon)^nrightarrow0$

But for the part two what I can see is that:
begin{align*}
limlimits_{nrightarrowinfty}intlimits_{0}^{1}f(x^n)dx &=limlimits_{nrightarrowinfty}limlimits_{epsilonrightarrow0}intlimits_{0}^{1-epsilon}f(x^n)dx
end{align*}
But there after if I need to make use of part (1) then I have to interchange the limits. So is it possible. If so, I would like to know what are the conditions we need to have for such an interchange.

Moreover I would like a feedback on the hint given. (Why is it given?)

Okay, so please let me state an analysis principle that will be very useful in numerous problems.

Interversion is painful.

Either there is a straightforward (« logically tautological », that is, that does not require any analysis) argument for doing it, or it requires thinking the whole reasoning through with more elaborate arguments (such as some form of uniform convergence).

So now, let $M$ be a bound for $f$. Let $epsilon >0$. I suggest that you prove that

$$left|int_0^1{f(x^n)}-f(0)right| leq left|int_0^{1-epsilon}{f(x^n)}-(1-epsilon)f(0)right| + Mepsilon + epsilon |f(0)|. $$

Thus for every $n$ large enough $$left|int_0^1{f(x^n)}-f(0)right| leq 3Mepsilon$$.

As $f$ is continuous at $0$, given $c>0$, there exists $delta>0$ such that $|f(x)-f(0)|<c$ whenever $|x|<delta$. So choose $n_0$ such that $(1-epsilon)^{n_0}<delta$. Then, for any $x<1-epsilon$ and $ngeq n_0$, we have $x^n<delta$; then $|f(x^n)-f(0)|<c$. Then
$$
(1-epsilon)(f(0)-c)=int_0^{(1-epsilon)}(f(0)-c),dxleqint_0^{1-epsilon}f(x^n),dxleq int_0^{(1-epsilon)}(f(0)+c),dx=(1-epsilon)(f(0)+c).
$$
As we can do this for all $c>0$, we get
$$
int_0^{1-epsilon}f(x^n),dx=(1-epsilon)f(0)
$$

For the second part, because $f$ is continuous and $[0,1]$ is compact, $f$ is bounded, say $aleq f(x)leq b$ for all $x$. Then, for any $epsilon>0$,
$$
int_0^1f(x^n),dx=int_0^{(1-epsilon)}f(x^n),dx+int_{(1-epsilon)}^1f(x^n),dx.
$$
For the second integral,
$$
epsilon, aleq int_{(1-epsilon)}^1f(x^n),dxleq epsilon, b.
$$
Then
$$
limsup_nint_0^1f(x^n),dx= (1-epsilon)f(0)+int_{(1-epsilon)}^1f(x^n),dxleq(1-epsilon)f(0)+epsilon,b.
$$
As this works for all $epsilon>0$,
$$tag1
limsup_nint_0^1f(x^n),dxleq f(0).
$$
Similarly,
$$
liminf_nint_0^1f(x^n),dx= (1-epsilon)f(0)+int_{(1-epsilon)}^1f(x^n),dxgeq(1-epsilon)f(0)+epsilon,a.
$$
As this works for all $epsilon>0$,
$$tag2
liminf_nint_0^1f(x^n),dxgeq f(0).
$$
Now $(1)$ and $(2)$ together imply that that $lim_nint_0^1f(x^n),dx$ exists and that
$$
lim_nint_0^1f(x^n),dx=f(0).
$$

So you managed to show that for each $varepsilonin(0,1)$ you get:
$$ limlimits_{ntoinfty}intlimits_0^{1-varepsilon}f(x^n),{rm d}x=(1-varepsilon)f(0) $$
So for each $varepsilonin(0,1)$ you can now look at the following:
$$ intlimits_0^1f(x^n),{rm d}x=intlimits_0^{1-varepsilon}f(x^n),{rm d}x+intlimits_{1-varepsilon}^1f(x^n),{rm d}x$$
Since you know that $f$ is bounded (a continuous function on a closed interval) you know there exists some $M>0$ such that:
$$ forall xin(0,1),-Mleq f(x)leq M $$
So you can show that for second interval you get:
$$ -Mvarepsilonleqintlimits_{1-varepsilon}^1f(x^n){rm d}xleq Mvarepsilon $$
Or in other words:
$$ intlimits_0^{1-varepsilon}f(x^n),{rm d}x-Mvarepsilonleq intlimits_0^1f(x^n),{rm d}xleq intlimits_0^{1-varepsilon}f(x^n){rm d}x+Mvarepsilon $$
Taking the limits on all sides as $ntoinfty$ you get using the first part:
$$(1-varepsilon)f(0)-Mvarepsilonleqlimlimits_{ntoinfty}intlimits_0^1f(x^n),{rm d}xleq(1-varepsilon)f(0)+Mvarepsilon $$
Lastly, taking the limit $varepsilonto 0$ gives you the result:
$$ limlimits_{ntoinfty}intlimits_0^1f(x^n),{rm d}x=f(0) $$

Another approach uses the mean value theorem for integrals. Indeed, set $g_n(x)=f(x^n).$ Then, there is a sequence $(c_n)subseteq (0,1-epsilon)$ such that $intlimits_0^{1-epsilon}g_n(x)dx=g_n(c_n)(1-epsilon)=f(c^n_n)(1-epsilon).$ Then, $c_n^nto 0$ as $nto infty$ and so $f(c_n^n)to f(0).$

For the second part, note that $int_0^1f(x^n),dx=int_0^{1-epsilon}f(x^n),dx+int_{1-epsilon}^1f(x^n),dx,$ so we just need to consider the second integral. But as $g_n$ is bounded, we can apply the dominated convergence theorem to write $lim int_{1-epsilon}^1f(x^n),dx=int_{1-epsilon}^1f(0)dx=epsilon f(0)$.

Therefore, $lim int_0^1f(x^n),dx= f(0).$