# Thread: integrable funtion and uniform convergence

1. ## integrable funtion and uniform convergence

Let $f_n$ be integrable on [0,1] for all $n$ and $f_n\rightarrow f$ uniformly. Show that $f$ is integrable and $\int f_n dx \rightarrow \int fdx$.

Since $f_n$ converges uniformly to $f$, for any $\epsilon >0$, there is $N$ such that for all $x$ and for all $n \geq N$, $|f-f_n|<\epsilon$. And since $|f|-|f_n| \leq|f-f_n|$, $\int |f|dx-\int |f_n|dx \leq \int |f|-|f_n|dx \leq \int |f-f_n|dx \rightarrow 0$.
So $\int |f| = lim \int |f_n|$.

i have a question here. i know that for each $n$, $\int |f_n| dx < \infty$ but how can i show that $lim \int |f_n|dx < \infty$.

help would be appreciated so much.

2. Originally Posted by PRLM
Let $f_n$ be integrable on [0,1] for all $n$ and $f_n\rightarrow f$ uniformly. Show that $f$ is integrable and $\int f_n dx \rightarrow \int fdx$.

Since $f_n$ converges uniformly to $f$, for any $\epsilon >0$, there is $N$ such that for all $x$ and for all $n \geq N$, $|f-f_n|<\epsilon$. And since $|f|-|f_n| \leq|f-f_n|$, $\int |f|dx-\int |f_n|dx \leq \int |f|-|f_n|dx \leq \int |f-f_n|dx \rightarrow 0$.
So $\int |f| = lim \int |f_n|$.

i have a question here. i know that for each $n$, $\int |f_n| dx < \infty$ but how can i show that $lim \int |f_n|dx < \infty$.

help would be appreciated so much.
can i assume that $f_n$ is bounded if $f_n$ is integrable?

3. I assume that you are working with the Riemann integral so then yes $f$ being integrable does mean that it is bounded.

4. To show integrability of $f$ you could use that $g:[a,b] \rightarrow \mathbb{R}$ is Riemann int. iff the set of discontinuities of $g$ has measure $0$ (ie. this set can be covered by a countable family of sets $(a_{i,\epsilon },b_{i,\epsilon })_{i\in \mathbb{N} }$ such that $\sum_{i=0}^{\infty } (b_{i, \epsilon }-a_{i, \epsilon }) < \epsilon$) .

5. Originally Posted by Jose27
To show integrability of $f$ you could use that $g:[a,b] \rightarrow \mathbb{R}$ is Riemann int. iff the set of discontinuities of $g$ has measure $0$ (ie. this set can be covered by a countable family of sets $(a_{i,\epsilon },b_{i,\epsilon })_{i\in \mathbb{N} }$ such that $\sum_{i=0}^{\infty } (b_{i, \epsilon }-a_{i, \epsilon }) < \epsilon$) .
I don't think this is true. Take f to be 0 on the rationals and 0 on the irrationals. Then the upper and lower Riemann sums don't agree for any partition of the interval (lets say [0,1]).

6. Originally Posted by putnam120
I don't think this is true. Take f to be 0 on the rationals and 0 on the irrationals. Then the upper and lower Riemann sums don't agree for any partition of the interval (lets say [0,1]).
I assume you meant Dirichlet's function instead of the zero constant. The problem is, Dirichlet's function is discontinous everywhere (ie. the measure of the set of discontinuities is positive)

7. Oh whoops sorry I misread what you meant. Then what about $f:[0,1]\to\{0,1\}$ where f is 1 on the Cantor set and 0 otherwise. Since the cantor set is uncountable you can't just cover it by intervals who's sum is small. Plus it has measure 0.

8. Originally Posted by putnam120
I assume that you are working with the Riemann integral so then yes $f$ being integrable does mean that it is bounded.
So if i can assume $f_n$ is bounded, i can find $M$ such that $|f_n| < M$ for all n, which mean $lim sup |f_n| < M$. So $\int |f|=\int lim|f_n|\leq\int limsup|f_n|<\infty$. is it correct?

9. Originally Posted by putnam120
Plus it has measure 0.
You answered yourself. To cover it the way I said remember how one constructed the set by taking intervlas of lenght $3^{-n}$ (if I rememeber correctly) and then intersecting all of them so it suffices to cover a "step" for sufficiently large $n$.

The fact that if such a cover exists then $f$ is integrable is more or less easy to see.

Well, I guess you also need to have f bounded on the hypothesis, but given it's Riemann it's always implicitly assumed.

10. Originally Posted by putnam120
Oh whoops sorry I misread what you meant. Then what about $f:[0,1]\to\{0,1\}$ where f is 1 on the Cantor set and 0 otherwise. Since the cantor set is uncountable you can't just cover it by intervals who's sum is small. Plus it has measure 0.
or are we supposed to used the assumption that $f_n\rightarrow f$ uniformly to say that $f_n$ and $f$ are both bounded? because if $f$ is not bounded i dont think $f_n$ can converge uniformly to $f$, can it? i dont think a uniform $\epsilon$ can work for all $x$ in the domain. would you give me your thoughts on this?

11. Originally Posted by PRLM
So if i can assume $f_n$ is bounded, i can find $M$ such that $|f_n| < M$ for all n, which mean $lim sup |f_n| < M$. So $\int |f|=\int lim|f_n|\leq\int limsup|f_n|<\infty$. is it correct?
No, it just means that for each n we have $f_n\le M_n$. So then since $f_n\to f$ uniformly we have that $|f-f_N|<\epsilon\Longrightarrow |f|\le |f_N|+\epsilon\le M_N+\epsilon$. Now take $M=\max(M_1,M_2,\dots,M_{N-1},M_N+\epsilon)$.

Thus $|f|\le M$ and $|f_n|\le M+\epsilon$. So even though what you said initially was correct (the $f_n$ are uniformly bounded) you were using the wrong definition for bounded.