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I'm stuck on the following exercise:

Let $f:\mathbb R\to \mathbb R$ be a nonnegative integrable function ($\int f<\infty$). Show that the function $$F(x)=\int_{-\infty}^xf$$ is continuous.

I was able to prove this very easily by concluding that $\lim_{x\to a} F(x) = F(a)$, but my textbook gave the following hint: "Use Theorem 10" which states:

THM 10:

Let $(f_n)$ be an increasing sequence of nonnegative measurable functions (with respect to the Lebesgue measure in the real line) and let $f=\lim f_n$ almost everywhere. Then $$\int f=\lim\int f_n$$

I just don't see how this theorem is related to this problem. How can we prove this using this theorem?

Eduardo Magalhães
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    Let $g$ be $f 1_{(-\infty, a)}$ and $g_n = f 1_{(-\infty, x_n)}$ with $x_n \rightarrow a$. Then apply Theorem 10. – David Gao Jan 11 '24 at 19:32
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    For a fixed $a \in \mathbb{R}$, consider a sequence of functions such as $$f_{n}(x) = \begin{cases}f(x), &x \leq a-\frac{1}{n} \ 0, &\text{otherwise}\end{cases}$$ – Prasiortle Jan 11 '24 at 19:32
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    @AnneBauval That isn't a duplicate at all. The other question isn't about measure theory and this one is. You can use the other question to solve this one, but that doesn't mean it's a duplicate. – David Lui Jan 11 '24 at 19:58
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    THM 10 as stated is false… take $$f_n(x) = n\cdot 1_{\left(0,\frac{1}{n}\right)}$$ then $(f_n)$ is a sequence of nonnegative measurable functions, $f_n \to f \equiv 0$ a.e. but $$\int f = 0 \not= 1 = \lim 1 = \lim \int f_n$$ – Gono Jan 11 '24 at 20:18
  • You are correct @Gono the sequence has to be increasing for the theorem to work. Already edited the post. Thx – Eduardo Magalhães Jan 11 '24 at 21:28
  • Alternatively, instead of monotone convergence, we could use dominated convergence, since the $g_n = f 1_{(-\infty, x_n)}$ are all dominated by the integrable function $|f|$. This will work even if $f$ is not necessarily nonnegative, merely integrable. – GEdgar Jan 11 '24 at 21:35
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    In my measure theory class, we are following the 3rd edition of "Real analysis" by H.L. Royden @AnneBauval – Eduardo Magalhães Jan 11 '24 at 23:13
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    @MathAttack I already suggested this link but David Lui's comment above finally convinced me to retract my vote to close as a duplicate, and delete my comment identical to yours. – Anne Bauval Jan 12 '24 at 10:23

1 Answers1

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Using Theorem 10 (Monotone convergence theorem), one can first easily prove that $F$ is left continuous: for every increasing convergent sequence $x_n\to x$, applying the theorem to $f_n:=1_{(-\infty, x_n)}f$, we find $F(x_n)\to F(x)$.

But thanks to the integrability of $f$, we can also prove that $F$ is right continuous: for every decreasing convergent sequence $x_n\to x$, applying the theorem to $f_n:=1_{(x_n,+\infty)}f=f-1_{(-\infty, x_n)}f$, we find $\left(\int f\right)-F(x_n)\to\left(\int f\right)-F(x)$, i.e. again, $F(x_n)\to F(x)$.

Anne Bauval
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