about $ lim_{n \to \infty } \int _0 ^ 1 f(nx) dx = lim_{n \to \infty } \int _0 ^ 1 f(n+x) dx = $ Let f be a continues function on $\Bbb{R} $  and   $lim_{n \to \infty }  \int _0 ^ 1 f(n+x) dx =a$
, prove :
$$ lim_{n \to \infty }  \int _0 ^ 1 f(nx) dx =a $$
My comment :
Let. $   \int f(x) dx =F(x)$ then 
$   \int f(n+x) dx =F(n+x), \int f(nx) dx = \frac {F(nx)}{n} $
$lim_{n \to \infty }  \int _0 ^ 1 f(n+x) dx =lim_{n \to \infty } F(n+1)-F(n) =a $ we must prove $ lim_{n \to \infty } \frac {F(n)}{n} =a $ 
In general case is following equation true  ?
$$ lim_{n \to \infty }  \int _0 ^ 1 f(nx) dx = lim_{n \to \infty }  \int _0 ^ 1 f(n+x) dx = $$
 A: That is a very good start.
You need to show that
$lim_{n \to \infty } F(n+1)-F(n) =a
$
implies
$lim_{n \to \infty } \frac {F(n)}{n} =a
$.
As I have described it before,
this is a standard
bad part/good part
limit theorem.
The good part is where
$F(n+1)-F(n)$
is close to $a$.
For any $\epsilon$,
we have
$|F(n+1)-F(n)-a)|
< \epsilon
$
for $n > N 
= N(\epsilon)
$.
The bad part is for
all smaller $n$,
where $F$ may misbehave.
For the good part,
for $n > N(\epsilon)$,
$-\epsilon
<F(n+1)-F(n)-a
< \epsilon
$
or
$a-\epsilon
<F(n+1)-F(n)
< a+\epsilon
$.
Summing this from
$N$ to $m-1$,
$(m-N)(a-\epsilon)
<F(m)-F(N)
<(m-N)(a+\epsilon)
$.
Now,
we let $m$ get large
and see what happens.
We have
$(m-N)(a-\epsilon)+F(N)
<F(m)
<(m-N)(a+\epsilon)+F(N)
$
so,
dividing by $m$,
$(1-\frac{N}{m})(a-\epsilon)+\frac{F(N)}{m}
<\frac{F(m)}{m}
<(1-\frac{N}{m})(a+\epsilon)+\frac{F(N)}{m}
$
or
$a-\epsilon-\frac{N}{m}(a-\epsilon)+\frac{F(N)}{m}
<\frac{F(m)}{m}
<a+\epsilon-\frac{N}{m}(a+\epsilon)+\frac{F(N)}{m}
$,
or, finally
$-\epsilon-\frac{N}{m}(a-\epsilon)+\frac{F(N)}{m}
<\frac{F(m)}{m}-a
<\epsilon-\frac{N}{m}(a+\epsilon)+\frac{F(N)}{m}
$.
By making $m$ large enough
(I will leave that to you),
we can make
$\frac{N}{m}(a-\epsilon)+\frac{F(N)}{m}
< \epsilon
$
and
$-\frac{N}{m}(a+\epsilon)+\frac{F(N)}{m}
< \epsilon
$.
This will make
$-2\epsilon
<\frac{F(m)}{m}-a
< 2\epsilon
$.
Therefore,
by first choosing
any $\epsilon > 0$
and then choosing
$m$ large anough
as a function of
$\epsilon$,
we can make
$|\frac{F(m)}{m}-a|
< 2\epsilon
$,
so that
$\lim_{m \to \infty} \frac{F(m)}{m}
=a
$.
