Prove that $ \sum_{i=1}^{\infty}\frac{f_{i}}{2^{i}} $ is integrable function.

Question

Let $ I\subseteq\mathbb{R}^{n} $ be a box.Let $ f_{i}:I\to[0,1] $ be integrable functions.

My first intuition was to use Weierstrass M-test, but we never proved it for multivariable functions and I cant see why would it hold.

The second intuition was to show that the set of discontinuities of $ \sum_{i=1}^{\infty}\frac{f_{i}}{2^{i}} $ is of measure zero, but im not sure how to show it.

Any help would be appreaciated. Thanks in advance.

How about Fubini's theorem on the product measure space $I \times \mathbb N$? The infinite sum $\sum\limits_{i=1}^{\infty}$ is the integral with respect to the counting measure on $\mathbb N$. — D_S, Feb 03 '21 at 19:30
The Weierstrass $M$-text does hold for multivariable functions, so you can use that approach. — Johnny El Curvas, Feb 03 '21 at 19:31
@DarthLubinus I see, that makes the solution easy. Although, we havent proved it in class and this is a question from one of the past exams, so if anyone can suggest another solution that would be great. — DirichletIsaPartyPooper, Feb 03 '21 at 19:32
while I would love to see a proof which avoids the $M$-test, I feel like any proof will essentially be proving (in a long-winded way) that the uniform limit of a sequence of Riemann-integrable functions is again Riemann-integrable. By the way, if you look again at the proof of the Weierstrass $M$-test; it doesn't matter what the domain of your function is; all that matters is that the target space is $\Bbb{R}$ (or $\Bbb{C}$ or really any complete normed vector space). The domain can be any arbitrary set non-empty $A$ (doesn't even have to be a subset of $\Bbb{R}^n$ or $\Bbb{C}$). — peek-a-boo, Feb 03 '21 at 19:40
@peek-a-boo I get what youre saying, problem is we havent proved that the uniform limit of a sequence of riemman integrable functions is riemann integrable for a metric space that is not $ \mathbb{R} $ or $ \mathbb{C} $ so Im uncomfortable with this solution — DirichletIsaPartyPooper, Feb 03 '21 at 19:46

score 2 · Accepted Answer · edited Feb 03 '21 at 20:11

Let $F_k:=\sum_{i=1}^k \frac{f_i}{2^i}$ and let $F=\lim_k F_k=\sum_{i=1}^\infty \frac{f_i}{2^i}$.

Since $f_i\ge 0$ for all $i$, the sequence $F_k$ is non-decrasing, so by Monotone Convergence Theorem $$\int_I F=\lim_k \int_I F_k.$$

But $f_i\le 1$ for all $i$, so (denoting $\lambda_n$ the $n$.dimensional Lebesgue measure) $$\int_I F_k=\sum_{i=1}^k\frac{1}{2^i}\int_If_i =\sum_{i=1}^k \frac{1}{2^i} \lambda_n(I)=\lambda_n(I)\sum_{i=1}^k\frac{1}{2^i}\overset{k\to\infty}{\longrightarrow}\lambda_n(I)<+\infty.$$ So, $F$ is integrable.

Prove that $ \sum_{i=1}^{\infty}\frac{f_{i}}{2^{i}} $ is integrable function.

1 Answers1