# Uniform limit of Lebesgue integrable functions on a space of infinite measure

Building off of this question: lebesgue integral uniform convergence

Given a measure space $(X, \Sigma, \mu )$, and a sequence $\{f_n\}_{n\in\mathbb{N}}$ of integrable functions $f_n:X\to \mathbb{R}$ that converge uniformly to some function $f$, we know that if $\mu(X)<\infty$ then $f$ is integrable, since we can bound $\int |f| \leq \int |f - f_n| + \int|f_n| \leq \mu(X)\epsilon + \int |f_n|$ for any $\epsilon >0$. Once we know $f$ is integrable we can use finiteness of $\mu(X)$ to show $\int f = \lim \int f_n$ (see linked question).

We have counter-examples in the case where $\mu(X) = \infty$ to show that $\int f \neq lim \int f_n$. Take for example $\{ \frac{\chi_{[0,n)}}{ n}\}_{n\in\mathbb{N}}$ (where $\chi$ is the characteristic function). This converges uniformly to $0$ but $\int 0 = 0 \neq 1 = \lim \int \frac{\chi_{[0,n)}}{n}$.

Question: do we have counterexamples to show that if $f_n \to f$ uniformly and $\mu(X) = \infty$, then $f$ is not necessarily integrable?

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I ask this question since I use that $\mu(X) < \infty$ in the proof that $f$ is integrable, but I cannot think of a counterexample as described above. – AnonymousCoward Feb 17 '12 at 7:30

Let $f_n: \mathbb{R}^+ \rightarrow \mathbb{R}$ be equal to $1/m$ on $(m-1,m]$ for $m=1, \ldots, n$ and $0$ elsewhere. It is clear that $f_n$ converges uniformly to the function that is equal to $1/m$ on $(m-1,m]$ for all $m \in \mathbb{N}$. This function is not integrable because $\sum_{i=1}^{\infty} 1/i$ diverges.