# Thread: The Limit of Integrals Is the Integral of the Limit

1. ## The Limit of Integrals Is the Integral of the Limit

I am asked to prove that, if $\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$ and $\displaystyle \lim_{n \rightarrow \infty} ||g_{n} - g|| = 0$, prove that $\displaystyle \lim_{n \rightarrow \infty} \int_{I} f_{n} \cdot g_{n} = \int_{I}f \cdot g.$

I doubt that I want to do this by brute force by providing some increasing sequence of upper functions that approach $f \cdot g$, though maybe I do since I know that $\displaystyle \lim_{n \rightarrow \infty} ||f_{n}|| = ||f||$ and likewise for $g$.

In any case, the alternative seems to be to use to the Dominated Convergence Theorem, in which case I want to show that $\{ f_{n} \cdot g_{n} \}$ converges almost everywhere (presumably to $f \cdot g$), and that there is a non-negative function which is integrable and dominates $|f_{n} \cdot g_{n}|$. But the first part seems like it might be false, since I don't know that the interval over which these functions are integrated is bounded.

Maybe I want to try an epsilon-delta argument?

2. What about uniform convergence? It seems that you are not given enough information though.

3. Nope, I'm not given that the functions are uniformly convergent, just that they are in L(I) and measurable.

4. Originally Posted by ragnar
I am asked to prove that, if $\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$ and $\displaystyle \lim_{n \rightarrow \infty} ||g_{n} - g|| = 0$, prove that $\displaystyle \lim_{n \rightarrow \infty} \int_{I} f_{n} \cdot g_{n} = \int_{I}f \cdot g.$

I doubt that I want to do this by brute force by providing some increasing sequence of upper functions that approach $f \cdot g$, though maybe I do since I know that $\displaystyle \lim_{n \rightarrow \infty} ||f_{n}|| = ||f||$ and likewise for $g$.

In any case, the alternative seems to be to use to the Dominated Convergence Theorem, in which case I want to show that $\{ f_{n} \cdot g_{n} \}$ converges almost everywhere (presumably to $f \cdot g$), and that there is a non-negative function which is integrable and dominates $|f_{n} \cdot g_{n}|$. But the first part seems like it might be false, since I don't know that the interval over which these functions are integrated is bounded.

Maybe I want to try an epsilon-delta argument?
Actually you only need that one of the sequences converges in norm, the other can converge weakly and the result still holds (assuming you're working in $L^2$, otherwise the products $f_ng_n$ and $fg$ may not be integrable):

$\displaystyle| \int_I f_ng_n - \int_I fg | \leq |\int_I f_ng_n - \int_I f_ng | +| \int_I f_ng - \int fg| \leq \| f_n\| \| g_n-g\| + | \int_I f_ng - \int_I fg | \rightarrow 0$.

Notice then that we only require $f_n$ to coverge weakly to $f$ and $g_n$ to converge in norm. The proof also shows that th result holds if $f_n,f \in L^p$ and $g_n,g\in L^q$ with $\frac{1}{p} + \frac{1}{q} =1$ and $1.