# Convergence Criterion for a Sequence of Square Lebesgue Integrable Functions

Recall from The Inner Product Space of Square Lebesgue Integrable Functions page that we saw that for any interval $I$, the set of square Lebesgue integrable functions on $I$, $L^2(I)$ is an inner product space if we define an inner product $(\cdot, \cdot)$ for all $f, g \in L^2(I)$ (where we regard two functions $f$ and $g$ to be the same function if $f(x) = g(x)$ almost everywhere on $I$) by:

(1)Furthermore, $L^2(I)$ becomes a normed space when we defined a norm $\| \cdot \|$ on $L^2(I)$ for all $f \in L^2(I)$ in terms of the inner product above by:

(2)We will now look at an important theorem regarding sequences of square Lebesgue integrable functions on an interval. We will show that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of square Lebesgue integrable functions on an interval where the sum of the norms of the functions in the sequence converges, then the sum of the functions in the sequence also converges - namely to a square Lebesgue integrable function on $I$.

Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of square Lebesgue integrable functions on an interval $I$ and suppose that $\displaystyle{\sum_{n=1}^{\infty} \| f_n(x) \|}$ converges. Then:a) $\displaystyle{\sum_{n=1}^{\infty} g_n}$ converges almost everywhere on $I$ to a limit function $f$ that is square Lebesgue integrable on $I$.b) $\displaystyle{\| f(x) \|= \lim_{n \to \infty} \biggr \| \sum_{k=1}^{n} f_k(x) \biggr \|}$. |

**Proof of a):**Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of square Lebesgue integrable functions on an interval $I$ and consider the series $\displaystyle{\sum_{n=1}^{\infty} \mid f_n(x) \mid}$. Let $(s_n(x))_{n=1}^{\infty}$ denote the sequence of partial sums to this series. Then for each $n \in \mathbb{N}$ we have that:

- Notice that for each $k$ that since $f_k$ is square Lebesgue integrable on $I$, $\mid f_k \mid$ is also square Lebesgue integrable on $I$.

*(This is because if $f_k$ is square Lebesgue integrable on $I$ then $f_k \in M(I)$ and $f_k^2 \in L(I)$. Since $f_k \in M(I)$ we have that there exists a sequence of step functions that converges to $f_k$. But then, the absolute value of each term in this sequence will be a sequence of step functions that converges to $\mid f_k \mid$, so $\mid f_k \mid \in M(I)$. Furthermore, since $f_k^2 \in L(I)$ we note that $\mid f_k \mid^2 = f_k^2 \in L(I)$. So $\mid f_k \mid \in M(I)$ and $\mid f_k \mid^2 \in L(I)$ and hence $\mid f_k \mid \in L^2(I)$).*

- So for each $n$, $s_n$ is a finite sum of square Lebesgue integrable functions on $I$ and by the theorems presented on the Linearity of Sums and Scalar Multiples of Square Lebesgue Integrable Functions we see that each $s_n$ is square Lebesgue integrable on $I$. Now since $(s_n(x))_{n=1}^{\infty}$ is an increasing sequence of functions, the new sequence of functions, $(s_n^2(x))_{n=1}^{\infty}$ is also increasing $(*)$. Furthermore, we're given that $\displaystyle{\sum_{n=1}^{\infty} \| f_n(x) \|}$ converges, say to some $M \in \mathbb{R}$, and the following chain of inequalities shows that the integral of $s_n^2(x)$ is bounded for each $n \in \mathbb{N}$:

- So $\left ( \int_I s_n^2(x) \: dx \right )_{n=1}^{\infty}$ is an increasing numerical sequence that is bounded above, so this sequence converges. So, since $(s_n^2(x))_{n=1}^{\infty}$ is a sequence of functions (that is actually Lebesgue integrable on $I$) that are increasing and $\displaystyle{\lim_{n \to \infty} \int_I s_n^2(x) \: dx}$ exists, we have by the Levi's Monotone Convergence Theorems that $(s_n^2(x))_{n=1}^{\infty}$ converges almost everywhere on $I$ to some function $F$ that is Lebesgue integrable on $I$ and additionally:

- But since $(s_n^2(x))_{n=1}^{\infty}$ converges almost everywhere on $I$ to some function $F \in L(I)$, we have that the sequence $(s_n(x))_{n=1}^{\infty}$ converges almost everywhere on $I$. Since $(s_n(x))_{n=1}^{\infty}$ is the sequence of partial sums to the series $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ we see that in fact $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges almost everywhere on $I$ to some Lebesgue integrable function, call it $f$.