Is the natural map $L^p(X) \otimes L^p(Y) \to L^p(X \times Y)$ injective? Let $X,Y$ be $\sigma$-finite measure spaces, and let $L^p(X) \otimes L^p(Y)$ be the algebraic tensor product.  The product has a natural map into $L^p(X \times Y)$ which takes $\sum a_{ij} f_i \otimes g_j$ to the function $F(x,y) = \sum a_{ij} f_i(x) g_j(y)$.  A moment's thought shows that this map is well-defined.  Is it also injective?
It seems that this should be true, but I can't see how to prove it.  Intuitively, one needs to show that if $\sum a_{ij} f_i(x) g_j(y) = 0$ a.e., then one should be able to cancel all the terms in the sum using bilinearity.  It is not quite clear how to do this without knowing anything about the terms.

EDIT: Here is a cleaned-up and corrected version of this answer, based on Pierre-Yves' suggestion (thanks!).  His answer above contains a much more complete version.
If $\sum_{i=1}^n a_{i} f_i \otimes g_i$ is not the zero element of $L^p(X) \otimes L^p(Y)$, we may assume without loss of generality that the $f_i$ are linearly independent.  We can also assume that $a_1 \ne 0$ and $g_1 \ne 0$.
Suppose that the corresponding function $F(x,y) = \sum_{i=1}^n a_{i} f_i(x) g_i(y) = 0$ a.e.  Since $g_1 \ne 0$, there is a measurable $B \subset Y$ of positive finite measure such that $\int_B g_1 \ne 0$ (the integral is finite by Hölder).    Then by Fubini's theorem, for a.e. $x$ we have
$$
0 = \int_{B} F(x,y)dy = \sum_{i=1}^n a_i \left(\int_{B} g_i\right) f_i(x).
$$
This contradicts the assumed linear independence of the $f_i$.
