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Let $(f_n)$ a sequence of continuous function that converge pointwise to $f$. We suppose that $f$ is Riemann integrable on $[a,b]$ and that $(f_n)$ converge uniformly on all compact $[a+\frac{1}{m},b-\frac{1}{m}]$. I want to show that $$\lim_{n\to \infty }\int_a^b f_n(x)dx=\int_a^b f(x)dx.$$

My proof

Let $c\in ]a,b[$. Then I'll try to show that $$\lim_{n\to \infty }\int_c^b f_n(x)dx=\int_c^b f(x)dx.$$

  • Step 1 I define $\displaystyle F_n(x)=\int_c^x f_n(t)dt$ for all $x\in [a,b]$ that is continuous on $[a,b]$. I also define $\displaystyle F(x)=\int_c^x f(x)dx$. By uniform convergence of $(f_n)_n$ to $f$ on all compact of $]a,b[$, we have that $f$ is continuous on all compact of $]a,b[$ and thus on $]a,b[$ and then so is $F$. Since $f$ is integrable, we can prolonge $F$ by continuity on $a$ and $b$, and thus $F$ is also continuous on $[a,b]$.

  • Step 2 What I want to prove is that $\displaystyle\lim_{n\to \infty }F_n(b)=F(b)$. I have that for all $n$ and for all $p$ big enough $$|F_n(b)-F(b)|\leq |F_n(b)-F_n(b-1/p)|+|F_n(b-1/p)-F(b-1/p)|+|F(b-1/p)-F(b)|.$$

Using continuity of $F_n$ and $F$, we have that $$|F_n(b)-F_n(b-1/p)|+|F(b-1/p)-F(b)|\underset{p\to \infty }{\longrightarrow } 0,$$ and thus $$|F_n(b)-F(b)|\leq |F_n(b-1/p)-F(b-1/p)|$$ for all $n$ and all $p$. Now my problem is the folowing one : For all $p$ $f_n\to f$ uniformly on $[c,b-1/p]$ and thus, $$|F_n(b-1/p)-F(b-1/p)|\underset{n\to \infty }{\longrightarrow }0$$ can I conclude that $$\lim_{n\to \infty }|F_n(b)-F(b)|=0\ \ ?$$

  • Step 3 The problem goes that way : if $\varepsilon>0$, by continuity of $F_n$ and $F$, there is $P$ s.t. $$|F_n(b)-F_n(b-1/P)|+|F(b-1/P)-F(b)|<\frac{2\varepsilon}{3},$$ and thus $$|F_n(b)-F(b)|<\frac{2\varepsilon}{3}+|F_n(b-1/P)-F(b-1/P)|.$$ Since $$\lim_{n\to \infty }F_n(b-1/P)=\lim_{n\to \infty }\int_c^{b-1/P}f_n(x)dx\underset{f_n\to f\ uniform}{=}\int_c^{b-1/P}f(x)dx=F(b),$$ and thus, there is $N$ s.t. $|F_n(b-1/P)-F(b-1/P)|<\frac{\varepsilon}{3}$ when $n\geq N$, and thus $|F_n(b)-F(b)|<\varepsilon$ when $n\geq N$, what prove the claim.

The thing is that here $P$ depend on $n$, no ? So it might be wrong.

Could someone tell me if it's correct or not ? And how to correct it if it's wrong ? The thing is that I have the impression that the step 2 is correct, but when I do it rigorously in step 3, it fail. What's the problem ?

1 Answers 1

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The result you're trying to prove is false. For example, on $[0,1]$ define $f_n(x) = n^2(1-x)x^n.$ Let $f(x) \equiv 0$ on $[0,1].$ Then $f_n \to f$ pointwise on $[0,1].$ And $f_n\to 0$ uniformly on each $[0,1-\epsilon].$ But

$$\int_0^1f_n = \frac{n^2}{n(n-1)}\to 1.$$

Therefore $\int_0^1f_n$ does not converge to $\int_0^1 f = 0.$

  • 0
    Thanks a lot for your answer. Indeed, my proof can't be correct. – 2017-02-05
  • 0
    If $(f_n)$ are uniformly bounded, the result work. But is my proof adaptable to this case ? – 2017-02-06