Show that $F:\mathbb{R}\rightarrow \mathbb{R}$, $F(x):=\int_{(-\infty,x]}fdm$ is continuous.

Question

Let $f:\mathbb{R}\rightarrow [0,\infty]$ be measurable with $\int_{\mathbb{R}}fdm<\infty$. Show that $F:\mathbb{R}\rightarrow \mathbb{R}$, $F(x):=\int_{(-\infty,x]}fdm$ is continuous.

Do I need to consider the case where $f$ is bounded?

Thanks in advance.

Answer 1

Since f is a nonnegative by definition of the Lebesgue integral $\forall \epsilon > 0$ there exists $h:\mathbb{R} \to \mathbb{R}$ such that h is bounded, measurable, nonnegative, finite support, $h \leq f$ $\forall x \in \mathbb{R}$, and $\int_{\mathbb{R}}{|f-h|} < \epsilon$.

WLOG assume $y<x$ Consider $|F(x)-F(y)|=|\int_{y}^{x}{f dm}| \leq \int_{y}^{x}{|f-h|} + \int_{y}^{x}{|h|} < \epsilon + \int_{y}^{x}{|h|}$. Now, h is bounded so $\exists M \in \mathbb{R}$ such that $|h| \leq M$ for all x. Select $\delta = \epsilon / M$.

Then you have that $\int_{y}^{x}{|h|} \leq M |x-y| < M \delta = \epsilon$. Hence, you have that $|F(x)-F(y)| < 2 \epsilon$ i.e is continuous.

Answer 2

Consequence of absolute continuity of Lebesgue integral.

Lemma Let $f : \mathbb{R}^d \to \mathbb{R}$ be a Lebesgue integrable function. Then for every $\epsilon>0$, $\exists \delta > 0$ such that $$ \mu(E) < \delta \implies \int_{E}|f|d\mu < \epsilon. $$

Let us first see how the lemma implies the result you are asking. $$ |F(x) - F(y)| = \left|\int_{(x,y]} fd\mu\right| \leq \int_{(x,y]}|f|d\mu<\epsilon $$ whenever $\mu((x,y]) = |y-x|<\delta$. We will now conclude by proving the lemma.

Proof of Lemma Let $E_n$ be a nested sequence of sets defined via: $$ E_n = \{x : |f(x)|\leq n\}. $$ Clearly, $0\leq \underbrace{|f|\mathcal{X}_{E_n}}_{=g_n}\nearrow |f|$. Thus by monotone convergence theorem, $$ \lim_{n\to\infty}\int g_n d\mu = \int fd\mu. $$ Take $N$ large enough so that $$ \int (f-g_N) d\mu < \frac{\epsilon}{2}. $$ Now, $$ \int_E f d\mu = \underbrace{\int_E (f-g_N) d\mu}_{<\frac{\epsilon}{2}} + \underbrace{\int_E g_N d\mu}_{\leq N \mu(E)}. $$ It suffices to pick $\delta$ so that $N\delta < \frac{\epsilon}{2}$. It proves the lemma. $\blacksquare$

Answer 3

Suppose $x_n \to x^+.$ Then

$$|F(x_n) - F(x)| = |\int_x^{x_n} f| \le \int_x^{x_n} |f| = \int_{\mathbb R}|f|\chi_{(x,x_n)}.$$

The integrands on the right $\to 0$ pointwise everywhere, and are bounded above on $\mathbb R$ by $|f|\in L^1(\mathbb R).$ By the DCT, $|F(x_n) - F(x)| \to 0.$ Thus $F$ is continuous from the right. The proof from the left is similar.