$underset{nrightarrow infty}{lim}int_{0}^{b_{n}}f_n(x)dx=int_0^1f(x)dx$












2












$begingroup$



Let $(f_n)_{ninmathbb{N}}$ ne a sequence of continuous functions on
$[0,1]rightarrowmathbb{R}$ that converges uniformly to
$f:[0,1]rightarrowmathbb{R}$. Let $(b_n)_{ninmathbb{R}}$ be an
increasing sequence of real numbers in $(0,1)$ that converges to 1.
Prove that : $$underset{nrightarrow infty}{lim}int_{0}^{b_{n}}f_n(x)dx=int_0^1f(x)dx$$




As far as I know if we have a sequence of functions converging uniformly to $f$ AND if they are uniformly bounded we can have the following: $$underset{nrightarrow infty}{lim}int_{0}^{1}f_n(x)dx=int_0^1f(x)dx$$.

But here it is not so. Even the uniform boundedness is also not mentioned.






analysis riemann-integration







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Uniform convergence is sufficient for the interchange of limit with the integral.
    $endgroup$
    – Math1000
    Jan 5 at 22:12










  • $begingroup$
    Thank you for correcting
    $endgroup$
    – DD90
    Jan 5 at 22:36
















2












$begingroup$



Let $(f_n)_{ninmathbb{N}}$ ne a sequence of continuous functions on
$[0,1]rightarrowmathbb{R}$ that converges uniformly to
$f:[0,1]rightarrowmathbb{R}$. Let $(b_n)_{ninmathbb{R}}$ be an
increasing sequence of real numbers in $(0,1)$ that converges to 1.
Prove that : $$underset{nrightarrow infty}{lim}int_{0}^{b_{n}}f_n(x)dx=int_0^1f(x)dx$$




As far as I know if we have a sequence of functions converging uniformly to $f$ AND if they are uniformly bounded we can have the following: $$underset{nrightarrow infty}{lim}int_{0}^{1}f_n(x)dx=int_0^1f(x)dx$$.

But here it is not so. Even the uniform boundedness is also not mentioned.






analysis riemann-integration







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Uniform convergence is sufficient for the interchange of limit with the integral.
    $endgroup$
    – Math1000
    Jan 5 at 22:12










  • $begingroup$
    Thank you for correcting
    $endgroup$
    – DD90
    Jan 5 at 22:36














2












2








2





$begingroup$



Let $(f_n)_{ninmathbb{N}}$ ne a sequence of continuous functions on
$[0,1]rightarrowmathbb{R}$ that converges uniformly to
$f:[0,1]rightarrowmathbb{R}$. Let $(b_n)_{ninmathbb{R}}$ be an
increasing sequence of real numbers in $(0,1)$ that converges to 1.
Prove that : $$underset{nrightarrow infty}{lim}int_{0}^{b_{n}}f_n(x)dx=int_0^1f(x)dx$$




As far as I know if we have a sequence of functions converging uniformly to $f$ AND if they are uniformly bounded we can have the following: $$underset{nrightarrow infty}{lim}int_{0}^{1}f_n(x)dx=int_0^1f(x)dx$$.

But here it is not so. Even the uniform boundedness is also not mentioned.






analysis riemann-integration







share|cite|improve this question









$endgroup$





Let $(f_n)_{ninmathbb{N}}$ ne a sequence of continuous functions on
$[0,1]rightarrowmathbb{R}$ that converges uniformly to
$f:[0,1]rightarrowmathbb{R}$. Let $(b_n)_{ninmathbb{R}}$ be an
increasing sequence of real numbers in $(0,1)$ that converges to 1.
Prove that : $$underset{nrightarrow infty}{lim}int_{0}^{b_{n}}f_n(x)dx=int_0^1f(x)dx$$




As far as I know if we have a sequence of functions converging uniformly to $f$ AND if they are uniformly bounded we can have the following: $$underset{nrightarrow infty}{lim}int_{0}^{1}f_n(x)dx=int_0^1f(x)dx$$.

But here it is not so. Even the uniform boundedness is also not mentioned.






analysis riemann-integration




analysis riemann-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 22:01









DD90DD90

2648




2648








  • 1




    $begingroup$
    Uniform convergence is sufficient for the interchange of limit with the integral.
    $endgroup$
    – Math1000
    Jan 5 at 22:12










  • $begingroup$
    Thank you for correcting
    $endgroup$
    – DD90
    Jan 5 at 22:36














  • 1




    $begingroup$
    Uniform convergence is sufficient for the interchange of limit with the integral.
    $endgroup$
    – Math1000
    Jan 5 at 22:12










  • $begingroup$
    Thank you for correcting
    $endgroup$
    – DD90
    Jan 5 at 22:36








1




1




$begingroup$
Uniform convergence is sufficient for the interchange of limit with the integral.
$endgroup$
– Math1000
Jan 5 at 22:12




$begingroup$
Uniform convergence is sufficient for the interchange of limit with the integral.
$endgroup$
– Math1000
Jan 5 at 22:12












$begingroup$
Thank you for correcting
$endgroup$
– DD90
Jan 5 at 22:36




$begingroup$
Thank you for correcting
$endgroup$
– DD90
Jan 5 at 22:36










1 Answer
1






active

oldest

votes


















4












$begingroup$

Hint:



For any $epsilon > 0$, we have for all sufficiently large $n$
$$left|int_0^{b_n} f_n - int_0^1 f right| leqslant int_0^{b_n}|f_n - f| + int_{b_n}^1 |f| leqslant epsilon b_n + sup_{x in [0,1]} |f(x)| (1 - b_n)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. It works. But I need a small clarification: is it always true that $|int f dx|leqint|f|dx$? If so can you please point out a website or a book which state this result? Thanks again!
    $endgroup$
    – DD90
    Jan 5 at 22:35






  • 1




    $begingroup$
    So $-|f(x)| leqslant f(x) leqslant |f(x)|$. One side gives us $int_0^1f leqslant int_0^1|f|$. The other side gives us $-int_0^1|f| leqslant int_0^1 f$. Thus $|int_0^1 f| leqslant int_0^1 |f|$.
    $endgroup$
    – RRL
    Jan 5 at 22:38












  • $begingroup$
    Ah! So it is the proof. Thank you!
    $endgroup$
    – DD90
    Jan 5 at 22:41






  • 1




    $begingroup$
    @DD90: You are welcome.
    $endgroup$
    – RRL
    Jan 5 at 22:42











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

Hint:



For any $epsilon > 0$, we have for all sufficiently large $n$
$$left|int_0^{b_n} f_n - int_0^1 f right| leqslant int_0^{b_n}|f_n - f| + int_{b_n}^1 |f| leqslant epsilon b_n + sup_{x in [0,1]} |f(x)| (1 - b_n)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. It works. But I need a small clarification: is it always true that $|int f dx|leqint|f|dx$? If so can you please point out a website or a book which state this result? Thanks again!
    $endgroup$
    – DD90
    Jan 5 at 22:35






  • 1




    $begingroup$
    So $-|f(x)| leqslant f(x) leqslant |f(x)|$. One side gives us $int_0^1f leqslant int_0^1|f|$. The other side gives us $-int_0^1|f| leqslant int_0^1 f$. Thus $|int_0^1 f| leqslant int_0^1 |f|$.
    $endgroup$
    – RRL
    Jan 5 at 22:38












  • $begingroup$
    Ah! So it is the proof. Thank you!
    $endgroup$
    – DD90
    Jan 5 at 22:41






  • 1




    $begingroup$
    @DD90: You are welcome.
    $endgroup$
    – RRL
    Jan 5 at 22:42
















4












$begingroup$

Hint:



For any $epsilon > 0$, we have for all sufficiently large $n$
$$left|int_0^{b_n} f_n - int_0^1 f right| leqslant int_0^{b_n}|f_n - f| + int_{b_n}^1 |f| leqslant epsilon b_n + sup_{x in [0,1]} |f(x)| (1 - b_n)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. It works. But I need a small clarification: is it always true that $|int f dx|leqint|f|dx$? If so can you please point out a website or a book which state this result? Thanks again!
    $endgroup$
    – DD90
    Jan 5 at 22:35






  • 1




    $begingroup$
    So $-|f(x)| leqslant f(x) leqslant |f(x)|$. One side gives us $int_0^1f leqslant int_0^1|f|$. The other side gives us $-int_0^1|f| leqslant int_0^1 f$. Thus $|int_0^1 f| leqslant int_0^1 |f|$.
    $endgroup$
    – RRL
    Jan 5 at 22:38












  • $begingroup$
    Ah! So it is the proof. Thank you!
    $endgroup$
    – DD90
    Jan 5 at 22:41






  • 1




    $begingroup$
    @DD90: You are welcome.
    $endgroup$
    – RRL
    Jan 5 at 22:42














4












4








4





$begingroup$

Hint:



For any $epsilon > 0$, we have for all sufficiently large $n$
$$left|int_0^{b_n} f_n - int_0^1 f right| leqslant int_0^{b_n}|f_n - f| + int_{b_n}^1 |f| leqslant epsilon b_n + sup_{x in [0,1]} |f(x)| (1 - b_n)$$






share|cite|improve this answer









$endgroup$



Hint:



For any $epsilon > 0$, we have for all sufficiently large $n$
$$left|int_0^{b_n} f_n - int_0^1 f right| leqslant int_0^{b_n}|f_n - f| + int_{b_n}^1 |f| leqslant epsilon b_n + sup_{x in [0,1]} |f(x)| (1 - b_n)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 22:17









RRLRRL

52.8k42573




52.8k42573












  • $begingroup$
    Thanks. It works. But I need a small clarification: is it always true that $|int f dx|leqint|f|dx$? If so can you please point out a website or a book which state this result? Thanks again!
    $endgroup$
    – DD90
    Jan 5 at 22:35






  • 1




    $begingroup$
    So $-|f(x)| leqslant f(x) leqslant |f(x)|$. One side gives us $int_0^1f leqslant int_0^1|f|$. The other side gives us $-int_0^1|f| leqslant int_0^1 f$. Thus $|int_0^1 f| leqslant int_0^1 |f|$.
    $endgroup$
    – RRL
    Jan 5 at 22:38












  • $begingroup$
    Ah! So it is the proof. Thank you!
    $endgroup$
    – DD90
    Jan 5 at 22:41






  • 1




    $begingroup$
    @DD90: You are welcome.
    $endgroup$
    – RRL
    Jan 5 at 22:42


















  • $begingroup$
    Thanks. It works. But I need a small clarification: is it always true that $|int f dx|leqint|f|dx$? If so can you please point out a website or a book which state this result? Thanks again!
    $endgroup$
    – DD90
    Jan 5 at 22:35






  • 1




    $begingroup$
    So $-|f(x)| leqslant f(x) leqslant |f(x)|$. One side gives us $int_0^1f leqslant int_0^1|f|$. The other side gives us $-int_0^1|f| leqslant int_0^1 f$. Thus $|int_0^1 f| leqslant int_0^1 |f|$.
    $endgroup$
    – RRL
    Jan 5 at 22:38












  • $begingroup$
    Ah! So it is the proof. Thank you!
    $endgroup$
    – DD90
    Jan 5 at 22:41






  • 1




    $begingroup$
    @DD90: You are welcome.
    $endgroup$
    – RRL
    Jan 5 at 22:42
















$begingroup$
Thanks. It works. But I need a small clarification: is it always true that $|int f dx|leqint|f|dx$? If so can you please point out a website or a book which state this result? Thanks again!
$endgroup$
– DD90
Jan 5 at 22:35




$begingroup$
Thanks. It works. But I need a small clarification: is it always true that $|int f dx|leqint|f|dx$? If so can you please point out a website or a book which state this result? Thanks again!
$endgroup$
– DD90
Jan 5 at 22:35




1




1




$begingroup$
So $-|f(x)| leqslant f(x) leqslant |f(x)|$. One side gives us $int_0^1f leqslant int_0^1|f|$. The other side gives us $-int_0^1|f| leqslant int_0^1 f$. Thus $|int_0^1 f| leqslant int_0^1 |f|$.
$endgroup$
– RRL
Jan 5 at 22:38






$begingroup$
So $-|f(x)| leqslant f(x) leqslant |f(x)|$. One side gives us $int_0^1f leqslant int_0^1|f|$. The other side gives us $-int_0^1|f| leqslant int_0^1 f$. Thus $|int_0^1 f| leqslant int_0^1 |f|$.
$endgroup$
– RRL
Jan 5 at 22:38














$begingroup$
Ah! So it is the proof. Thank you!
$endgroup$
– DD90
Jan 5 at 22:41




$begingroup$
Ah! So it is the proof. Thank you!
$endgroup$
– DD90
Jan 5 at 22:41




1




1




$begingroup$
@DD90: You are welcome.
$endgroup$
– RRL
Jan 5 at 22:42




$begingroup$
@DD90: You are welcome.
$endgroup$
– RRL
Jan 5 at 22:42


















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