Primitive of non $L^1$ function is continuous

Question

I'm beginning with Lebesgue integration this semestre and still quite confused with it. In my course notes, the Lebesgue integration is defined on all measurable functions $R^d \to C$ (infinite integrations are accepted). A function is integrable if $\int |f|$ is finite. The set of all integrable functions in $R^d$ is denoted by $L^{1}(R^d)$.

There is an exercice in the notes is to prove that the function $g: x \mapsto \int_{a}^{x} f dt$ where $f \in L^{1}(R^d)$ is continous.

But I can prove that even if $f \notin L^{1}$, it only needs to be measurable, then $g$ is continuous. Actually, $$ |g(t+h) - g(t)| = |\int_{t}^{t+h} f dx| \leq \int (1_{[t,t+h]} \times |f|) dx $$ Let's consider the monotone sequence of functions $f_{n} = 1_{[t,t+\frac{1}{n}]} \times |f|$ which point-wise converges to $1_{[t,t]} \times |f|$, note that $\int f_{n} dx= \int_{t}^{t+\frac{1}{n}} |f| dx $. Applying the Beppo-Levi theorem: $$ lim_{n\to \infty} \int f_{n} dx = \int lim_{n \to \infty} f_{n} $$ or $$ lim_{n\to \infty} \int_{t}^{t+\frac{1}{n}} |f| dx= \int_{t}^{t} |f| dx = 0 $$

then $g(t)$ is continuous for any $t$.

Of course this proof cannot be correct since a counter-example is $f = \frac{1}{x}$ which is not continous at $0$, but I still cannot find where is the error (is the Beppo-Levi theorem applicable for finite integration only?)

Answer 1

You can also think in the measure $\phi(E) = \int_E |f| dx$ and consider the decreasing measurable sets $E_n = [t,t+1/n]$ as you defined them. Of course the intersection of all of them is $\{t\}$, therfore $$0 = \int_t^t |f| = \phi(\{t\})=\phi(\cap E_i) = lim_{i\rightarrow \infty} \phi(E_i)$$ The last equality is true if $\phi(E_1)$ is finite (or it is finite for some $i$), there is when we had into account the integrability of $f$, otherwise the equality may fails.

Answer 2

The problem with the reasoning is Beppo-Levi's theorem (AKA monotone convergence theorem) is more restrictive when applied to decreasing rather than increasing sequences. In the decreasing case, one has to assume that one (and hence every) $f_n$ is integrable. (See the corollary here, for instance.)

Another vivid illustration of the same issue is with the decreasing sequence of indicator functions $$f_n=\chi_{[-1/n,1/n]\times\mathbb R}:\mathbb R^2\to\mathbb R$$ which converge to zero pointwise.