Anti-Aliasing and the Fourier Transform, Gonzalez Digital Image Processing In Gonzalez book Digital Image Processing, section 4.34 (third edition), he writes:

Unfortunately, except for some special cases mentioned blow, aliasing
is always present in sampled signals because, even if the original
sampled function is band-limited, infinite frequency components are
introduced the moment we limit the duration of the function, which we
always have to do in practice.
For example, suppose that we want to limit the duration of a
band-limited function $f(t)$ (i.e. a function whose Fourier transform
is non-zero only on a closed interval of range of frequencies), to an
interval say $[0, T]$. We can do this by multiplying $f(t)$ by the
function
$h(t)= 1 $ if $t \in [0,T]$, and is $0$ otherwise.
Then from the convolution theorem we know that the transform of this
product $h(t)f(t)$ is the convolution of the transforms of the
functions. Even if the transform of $f(t)$ is band-limited, convolving
it with $F(h(t))=H(\mu)$ will yield a result with frequency components
that are infinite.

This very last statement is what I am not sure about. If the Fourier transform of $f$ is band-limited, then outside of a closed interval, the transformed function will be $0$, and so I am not sure for which frequency components the convolution of the transforms will have infinite frequency. Any insights appreciated.
Note that in order to obtain the Fourier transform of the windowed time domain signal $f(t)h(t)$, you need to convolve the Fourier transforms of $f(t)$ and $h(t)$. We know that the Fourier transform of $f(t)$ is zero outside some interval (because $f(t)$ is band-limited). However, since $h(t)$ is time-limited, i.e., it is zero outside some time interval, we know that its Fourier transform cannot be band-limited. For this specific example of a rectangular window $h(t)$, we know that the corresponding Fourier transform is a sinc function. Convolving any function with a sinc function results in a function that extends from $-\infty$ to $+\infty$. Consequently, the Fourier transform of the windowed signal $f(t)h(t)$ also extends from $-\infty$ to $+\infty$. Consequently, the signal $f(t)h(t)$ cannot be band-limited.

In sum, a band-limited signal cannot be time-limited, and a time-limited signal cannot be band-limited. Note, however, that you cannot conclude that a signal is band-limited if it is not time-limited and vice versa. There are signal that are neither time-limited nor band-limited.