If an integral operator $T : L^p(\Bbb{R}) \to L^q(\Bbb{R})$ is well-defined, it is bounded 
Let $p,q \in [1,+\infty\rangle$ and let $K : \Bbb{R}^2 \to \Bbb{R}$ be a measurable function such that the linear map $T : L^p(\Bbb{R}) \to L^q(\Bbb{R})$ given by$$(Tf)(x) = \int_{\Bbb{R}} K(x,y)f(y)\,dy, \quad \text{for a.e. $x \in \Bbb{R}$ and $f \in L^p(\Bbb{R})$}$$
is well-defined. Can we conclude that $T$ is bounded?

My attempt:
We shall use the Closed Graph Theorem to show that $T$ is bounded.
Since $T$ is well-defined, for every $f \in L^p(\Bbb{R})$ the function $Tf$ is a well-defined $L^q(\Bbb{R})$ function which is equal to the above integral for a.e. $x \in \Bbb{R}$. In particular, the integral $\int_{\Bbb{R}} K(x,\cdot)f(\cdot)$ exists for a.e. $x \in \Bbb{R}$.
We can conclude that for every $f \in L^p(\Bbb{R})$ there exists a set $N_f \subseteq \Bbb{R}$ of measure zero such that for all $x \in \Bbb{R}\setminus N_f$ holds $K(x,\cdot)f(\cdot) \in L^1(\Bbb{R})$.
Here comes the unclear part. Can we say that there exists a "global" set $N \subseteq \Bbb{R}$ measure zero such that for every $x \in \Bbb{R}\setminus N$ we have
$$K(x,\cdot)f(\cdot) \in L^1(\Bbb{R}),\quad \text{ for all }f \in L^1(\Bbb{R})?$$
If so, we can proceed as in this answer:
For all $x \in \Bbb{R}\setminus N$ we define a linear map $A_x : L^p(\Bbb{R}) \to L^1(\Bbb{R})$ as $A_x(f) := K(x,\cdot)f(\cdot)$. It is easy to show that $A_x$ is bounded by the Closed Graph Theorem.
Now for all $x \in \Bbb{R}\setminus N$ we define the linear functional $l_x : L^p(\Bbb{R}) \to \Bbb{C}$ as $$l_x(f) := \int_{\Bbb{R}} K(x,\cdot)f(\cdot) = \int_{\Bbb{R}} A_x(f) = (Tf)(x), \quad f\in L^p(\Bbb{R})$$
Since $A_x$ is bounded, $l_x$ is bounded as well.
Now assume $f_n \xrightarrow{L^p} 0$ and $Tf_n \xrightarrow{L^q} f \in L^q(\Bbb{R})$ and we wish to show that $f = 0$. Since $l_x$ is continuous for all $x \in \Bbb{R}\setminus N$, for all such $x$ we have
$$(Tf_n)(x) = l_x(f_n) \xrightarrow{n\to\infty} l_x(0) = 0$$
and hence $Tf_n \xrightarrow{\mathrm{a.e.}} 0$. From $Tf_n \xrightarrow{L^q} f$ by passing to a subsequence we conclude $f = 0$. Hence, $T$ is bounded by CGT.
So basically, my question boils down to the fact whether the quantifiers "for every" and "for almost every" commute. Are the statements

*

*$$(\text{for every } f \in L^p(\Bbb{R}))(\text{for a.e. }x \in \Bbb{R}) \quad K(x,\cdot)f(\cdot)\in L^1(\Bbb{R})$$

*$$(\text{for a.e. }x \in \Bbb{R})(\text{for every } f \in L^p(\Bbb{R})) \quad K(x,\cdot)f(\cdot)\in L^1(\Bbb{R})$$
equivalent?
 A: The equivalence is false. In fact, if the second assertion holds for every $f\in L^p(\mathbb R)$ it follows that $K(x,\cdot)\in L_{p'}(\mathbb R)$ with $\frac1p+\frac1{p'}=1$ which is not true, in general if $T\colon L^p\to L^q$.
A more concrete counterexample is $K(x,y)=g(x-y)$ with $g\in L^1(\mathbb R)$ where the second assertion is false, in general, but the first holds with $p=1$. (In fact, you can pick $q=1$ by Fubini-Tonelli.)
However, the operator $T$ is automatically bounded - you can find the proof in Banach's classical monograph, BTW. The proof goes as follows:
Let $f_n\to0$ in $L^p$ and $Tf_n\to g$ in $L^q$. As you observe, one has to show that $g=0$. Passing to a subsequence, you can assume that $Tf_n\to g$ almost everywhere. Passing to a further subsequence, you can assume that $f_n\to0$ a.e., and moreover, that there is a function $h\in L^p$ such that $\lvert f_n(y)\rvert\le h(y)$ for almost every $y$. (The latter assertion is a little bit tricky to show and is usually proved implicitly in the proof of the completeness of $L^p$; in Rudin's "Real and complex analysis" you will find the proof.)
Now it suffices to observe that for the particular function $h$ the function $K(x,\cdot)h(\cdot)$ is integrable for almost all $x$. Hence, for almost all $x$ you can use Lebesgue's dominated convergence theorem to show that $Tf_n(x)\to0$.
A: This is just to provide a few more details to the answer provided ny @MartinVäth.

*

*Assumptions: $(X,\mathscr{F},\mu)$ and $(Y,\mathscr{G},\nu)$ are $\sigma$--finite measure spaces, and $p,q\geq1$; $K:(X\times Y,\mathscr{F}\otimes\mathscr{G})\rightarrow\mathbb{R}$ $$Tf(x):=\int_Yf(y)K(x,y)\,\nu(dy)\in L_q(\mu)$$
whenever $f\in L_p(\nu)$.


*As suggested by the OP, we apply the closed graph theorem to show that $T:L_p(\nu)\rightarrow T_q(\mu)$ is bounded.


*It suffices to show that for any sequence $\{f_n:n\in\mathbb{N}\}\subset L_p(\nu)$ such that $(f_n,Tf_n)\xrightarrow{n\rightarrow\infty}(0,g)$ in $L_p(\nu)\times L_q(\mu)$, we have that $g=0$.


*Claim:  There is a set $A\subset Y$ with $\nu(A)=0$ and $h\in L_p(\nu)$ such that on $X\setminus A$ and along a subsequence $n'$ $|f_n'|\leq h$ and $f_{n'}\xrightarrow{n\rightarrow\infty}0$.
Proof: Take a subsequence  $n_k$ such that $\|f_n-f_{n_k}\|_p<2^{-k}$ for all $n\geq n_k$, and set $g=\sum_k|g_{n_k}-g_{n_{k+1}}|$. Then $0\leq g<\infty$ $\nu$-a.s., and $f'=f_{n_1}+\sum^\infty_{j=1}f_{n_{j+1}}-f_{n_j}$ converges absolutely $\nu$-a.s. For $n\geq n_k$
$$\|f'-f_n\|_p\leq\|f'-f_{n_k}\|_p+\|f_n-f_{n_k}\|_p\leq \Big\|\sum^\infty_{j=k}(f_{n_{j+1}}-f_{n_j})\Big\|_p+2^{-k}\leq 2^{-k+1}\xrightarrow{k\rightarrow\infty}0$$
Hence $f=f'$  and $|f_{n_k}|\leq|f|+g=h$ $\nu$-a.s. Let $A$ be the exceptional set in $Y$. 
By a similar argument, there is subsequence $k'$  along which
$$Tf_{n_{k'}}\xrightarrow{k'\rightarrow\infty}g\qquad\text{$\mu$--a.s.}$$
Since
$$\int_X|Th(x)|^q\,\mu(dx)=\int_X\Big|\int_YK(x,y) h(y)\,\nu(dy)\Big|^q\,\mu(dx)<\infty$$
there is a set $B\subset X$  with $\mu(B)=0$ such that $|K(x,\cdot)\, h(\cdot)|\in L_1(\nu)$ for all $x\in X\setminus B$. On $(X\setminus B)\times(Y\setminus A)$ we have
$$|K(x,y) f_n(x)|\leq |K(x,y) h(x)|$$
and
$$K(x,y)f_n(x)\xrightarrow{n\rightarrow\infty}0$$
By dominated convergence, for all $y\in X\setminus B$,
$$\lim_n\int_X K(x,y) f_n(x)\,d\mu(dx)=0$$
and
$$|\int_XK(x,y) f_n(x)\,d\mu(dx)|\leq\int_X|K(x,y) h(x)|\,\mu(dx)\in L_q(\nu)$$
Another application of dominated convergence implies that
$$
  \lim_n\int_Y\Big|\int_XK(x,y)\,f(x)\,\mu(dx)\Big|^p\,\mu(dy)=0
  $$
That is, $Kf_n\xrightarrow{n\rightarrow\infty}0$ in $L_q(\nu)$. This implies that $g=0$ $\mu$-a.s. The closed graph theorem implies that $T$ is a bounded operator, i.e., there is $c>0$ such that $\|Tf\|_{L_q(\nu)}\leq c\|f\|_{L_p(\mu)}$ for all $f\in L_p(\mu)$. $\Box$
