Proving that $F$ is continuous

Question

Let $f:[0,1] \times \mathbb{R} \to \mathbb{R}$ continuous function and consider $F(y)= \int_0^1 f(x,y) dm(x)$. I want to show that $F$ is continuous.

I tried: Take $y' \in \mathbb{R}$. Let $(y_n)_n$ any sequence such that $y_n \to y'$. I have to show that $\lim F(y_n)=F(y')$. For that I want to use some convergence theorem. Define $f_n(x)=f(x,y_n)$ using continuity of $f$ then:

$$ \lim_{n \to \infty} f_n(x)= \lim_{n \to \infty} f(x,y_n)= f(x,y)$$

I cannot finish. I cannot the limit of the sequence $f_n$.

Answer 1

Fix $y \in \mathbb{R}$. Note that $f$ is uniformly continuous in the compact set $K:= [0,1] \times [y-1,y+1]$. Now, let $\varepsilon > 0$.

By uniform continuity in $K$, there exists $\delta > 0$ such that if $\|(x,y)-(x',y')\| < \delta$ and $(x,y),(x',y') \in K$ then $|f(x,y)-f(x',y')| < \varepsilon/2$. Without loss of generality, we can additionally assume $\delta < 1$.

If $|y-y'| < \delta$, then we get that $\|(x,y)-(x,y')\| < \delta$ for each $x \in [0,1]$. Moreover it has to be $(x,y),(x,y') \in K$ because $|y-y'| < \delta < 1$. This says that

$$ |f(x,y)-f(x,y')| < \varepsilon/2 \quad (\forall y' \in B_\delta(y), x \in [0,1]) $$

and thus

$$ |F(y)-F(y')| \leq \int_0^1|f(x,y)-f(x,y')|d\mu \leq \int_0^1\varepsilon/2 \cdot d\mu = \varepsilon/2 < \varepsilon. $$

which concludes the proof: for a given $\varepsilon$, we have found $\delta > 0$ such that if $|y-y'| < \delta$, then $|F(y)-F(y')| < \varepsilon$.

Answer 2

If we are allowed to use Lebesgue integration theory, it is easy.

Let $a\in\mathbb{R}$ be arbitrary. We go to show that $F$ is continuous at $a$. Since $f$ is continuous on the compact set $[0,1]\times[a-1,a+1]$, there exists $M>0$ such that $|f(x,y)|\leq M$ for any $(x,y)\in[0,1]\times[a-1,a+1]$. Let $(y_{n})$ be an arbitrary sequence that converges to $a$. Without loss of generality, we may assume that $y_{n}\in[a-1,a+1]$ for all $n$. (For clarity reason for beginners...) For each $n$, define a function $g_{n}:[0,1]\rightarrow\mathbb{R}$ by $g_{n}(x)=f(x,y_{n})$. Define $g:[0,1]\rightarrow\mathbb{R}$ by $g(x)=f(x,a)$. Clearly $g_{n}$ and $g$ are Lebesgue integrable (in fact, they are continuous). Moreover, $|g_{n}|\leq M$ and $g_{n}\rightarrow g$ pointwisely by the continuity of $f$ and the fact that $(x,y_{n})\rightarrow(x,a)$. Therefore \begin{eqnarray*} & & \lim_{n\rightarrow\infty}F(y_{n})\\ & = & \lim_{n\rightarrow\infty}\int_{0}^{1}g_{n}(x)dx\\ & = & \int_{0}^{1}g(x)dx\\ & = & F(a). \end{eqnarray*} By Heine's Theorem, it follows that $F$ is continuous at $a$.

Answer 3

Another proof that does not involve any Lebesgue integration theory:

Let $b\in\mathbb{R}$ be arbitrary. We go to prove that $F$ is continuous at $b$. Let $\varepsilon>0$ be given. For each $a\in[0,1]$, since $f$ is continuous at $(a,b)$, there exists $\delta_{a}>0$ such that $|f(x,y)-f(a,b)|<\varepsilon$ whenever $|x-a|<\delta_{a}$ and $|y-b|<\delta_{a}$. Note that $\{(a-\delta_{a},a+\delta_{a})\mid a\in[0,1]\}$ is an open covering of the compact set $[0,1]$, so there exists a finite sub-cover. Choose a finite sub-cover and denote it by $\{(a_{i}-\delta_{a_{i}},a_{i}+\delta_{a_{i}})\mid i=1,\ldots,n\}$ for some $n\in\mathbb{N}$ and $a_{i}\in[0,1]$. Define $\delta=\frac{1}{2}\min_{1\leq i\leq n}\delta_{a_{i}}>0$. We go to show that $|F(y)-F(b)|\leq2\varepsilon$ whenever $|y-b|<\delta$.

Let $y\in\mathbb{R}$ such that $|y-b|<\delta$. Let $x\in[0,1]$ be arbitrary, then $x\in(a_{i}-\delta_{a_{i}},a_{i}+\delta_{a_{i}})$ for some $i$. For the point $(x,y)$, we clearly have $|x-a_{i}|<\delta_{a_{i}}$ and $|y-b|<\delta<\delta_{a_{i}}$. Therefore $|f(x,y)-f(a_{i},b)|<\varepsilon$. For the point $(x,b),$we also have $|x-a_{i}|<\delta_{a_{i}}$ and $|b-b|=0<\delta_{a_{i}}$, so $|f(x,b)-f(a_{i},b)|<\varepsilon$. It follows that \begin{eqnarray*} |f(x,y)-f(x,b)| & \leq & |f(x,y)-f(a_{i},b)|+|f(a_{i},b)-f(x,b)|\\ & < & 2\varepsilon. \end{eqnarray*} It follows that \begin{eqnarray*} & & |F(y)-f(b)|\\ & \leq & \int_{0}^{1}|f(x,y)-f(x,b)|dx\\ & \leq & 2\varepsilon. \end{eqnarray*}

Answer 4

Suppose $y_n \to y$, note that $f$ is bounded by some $B$ on the compact set $[0,1] \times \{y_n\}_n \cup \{y\}$, hence $|f(x,y_n) | \le B$, which is integrable. Since $f(x,y_n) \to f(x,y)$ we see that $F(y_n) \to F(y)$.